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# NCERT Solutions for Class 12 Maths Chapter 7 - In Hindi

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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals In Hindi pdf download

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Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Mathematics Chapter ७ – समाकलन

### प्रश्नावली 7.1

निम्नलिखित फलनों के प्रतियावकलज ( समाकलन ) निरीक्षण विधि द्वारा ज्ञात कीजिए।

1. $\mathbf{{\text{sin2x}}}$

उत्तर:  $\mathbf{\int {{\text{sin2x}}\;{\text{dx}}} }$

जेसे कि हम जानते है –

$\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cos2x}}\;{\text{ = }}\;{\text{ - 2sin2x}}$

$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cos2x}}} \right)\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{( - sin2x)2}}\;{\text{ = }}\;{\text{sin2x}}$

$\int {{\text{sin}}} {\text{2xdx}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cos2x + C}}$

2. $\mathbf{{\text{cos3x }}}$

उत्तर: $\int {{\text{cos}}\;{\text{3x}}}$

जेसे कि हम जानते है –

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sin3x}}\;{\text{ = }}\;{\text{ - 3cos3x}} \hfill \\ \begin{array}{*{20}{l}} {\text{ }} \\ {\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{3}}}{\text{sin3x}}} \right)\;{\text{ = }}\;{\text{cos3x}}} \\ {\int {{\text{cos}}} {\text{3xdx}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{\text{3}}}{\text{sin3x + C}}} \end{array} \hfill \\ \end{align}

3. $\mathbf{{{\text{e}}^{^{{\text{2x}}}}}}$

उत्तर: $\int {{{\text{e}}^{^{{\text{2x}}}}}\;dx}$

जेसे कि हम जानते है –

\begin{align} \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{2}}^{{\text{2x}}}}} \right)\;{\text{ = }}\;{\text{2}}{{\text{e}}^{{\text{2x}}}} \hfill \\ \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{{\text{2x}}}}} \right)\;{\text{ = }}\;{{\text{e}}^{{\text{2x}}}} \hfill \\ \int {{{\text{e}}^{{\text{2x}}}}} {\text{dx}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{{\text{2x}}}}{\text{ + C}} \hfill \\ \end{align}

4. $\mathbf{{{\text{(ax + b)}}^{\text{2}}}}$

उत्तर: $\int {{{{\text{(ax + b)}}}^{\text{2}}}\;{\text{dx}}}$

जेसे कि हम जानते है –

$\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{(ax + b)}}^{\text{3}}}\;{\text{ = }}\;{\text{2}}{{\text{e}}^{{\text{2x}}}}$

$\dfrac{{\text{d}}}{{{\text{dx}}}}\dfrac{{\text{1}}}{{{\text{3a}}}}{{\text{(ax + b)}}^{\text{3}}}\;{\text{ = }}\;{{\text{(ax + b)}}^{\text{2}}}$

$\int {{{{\text{(ax + b)}}}^{\text{2}}}} {\text{dx}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{3a}}}}{{\text{(ax + b)}}^{\text{3}}}{\text{ + C}}$

5. $\mathbf{{\text{sin2x - 4}}{{\text{e}}^{{\text{3x}}}}}$

उत्तर: $\int {\left( {{\text{sin2x - 4}}{{\text{e}}^{{\text{3x}}}}} \right)} {\text{dx}}\;$

$\int {\left( {{\text{sin2x - 4}}{{\text{e}}^{{\text{3x}}}}} \right)} {\text{dx}}\;{\text{ = }}\;\int {{\text{sin}}} {\text{2xdx - 4}}\int {{{\text{e}}^{{\text{3x}}}}}$   …(1)

$\int {{\text{sin}}\;{\text{2x}}\;{\text{dx}}\;{\text{ = }}\;{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cos}}\;{\text{2x}}}$   ….(2)

$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{e}}^{{\text{3x}}}}} \right)\;{\text{ = 3}}{{\text{e}}^{{\text{3x}}}}\;{\text{ = }}\int {{{\text{e}}^{{\text{3x}}}}} {\text{dx = }}\dfrac{{\text{1}}}{{\text{3}}}{{\text{e}}^{{\text{3x}}}}$   …(3)

समीकरण (1),(2),(3) से

$\ \int {\left( {{\text{sin2x - 4}}{{\text{e}}^{{\text{3x}}}}} \right)} {\text{dx}}\;{\text{ = }}\;\;{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cos}}\;{\text{2x - }}\dfrac{{\text{4}}}{{\text{3}}}{{\text{e}}^{{\text{3x}}}}{\text{ + c}}$

6. $\mathbf{\int {{\text{(4}}{{\text{e}}^{{\text{3x}}}}{\text{ + 1)dx}}} }$

उत्तर: $\int {\left( {{\text{4}}{{\text{e}}^{{\text{3x}}}}{\text{ + 1}}} \right)} {\text{dx}}\;$

\begin{align} \int {\left( {{\text{4}}{{\text{e}}^{{\text{3x}}}}{\text{ + 1}}} \right)} {\text{dx}}\;{\text{ = }}\;{\text{4}}\int {{{\text{e}}^{{\text{3x}}}}} {\text{dx + }}\int {\text{1}} {\text{dx}} \hfill \\ {\text{4}} \times \dfrac{{{{\text{e}}^{{\text{3x}}}}}}{{\text{3}}}{\text{ + x + C}}\;{\text{ = }}\;\dfrac{{\text{4}}}{{\text{3}}}{{\text{e}}^{{\text{3x}}}}{\text{ + x + C}} \hfill \\ \end{align}

7.$\mathbf{\int {{{\text{x}}^{\text{2}}}{\text{(1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{)dx}}} }$

उत्तर: $\int {{{\text{x}}^{\text{2}}}} \left( {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right){\text{dx}}\;$

\begin{align} \int {{{\text{x}}^{\text{2}}}} \left( {{\text{1 - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right){\text{dx}}\;{\text{ = }}\;\int {{{\text{x}}^{\text{2}}}} {\text{dx - }}\int {\text{1}} {\text{dx}} \hfill \\ \dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{\text{2 + 1}}}}{\text{ - x + C}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ - x + C}} \hfill \\ \end{align}

8. $\mathbf{\int {{\text{(a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c)dx}}} }$

उत्तर: $\int {\left( {{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c}}} \right)} {\text{dx}}\;$

$\int {\left( {{\text{a}}{{\text{x}}^{\text{2}}}{\text{ + bx + c}}} \right)} {\text{dx}}\;{\text{ = }}\;{\text{a}}\int {{{\text{x}}^{\text{2}}}} {\text{dx + c}}\int {\text{1}} {\text{dx}}\;$

${\text{ = }}\;{\text{a}}\dfrac{{{{\text{x}}^{{\text{2 + 1}}}}}}{{{\text{2 + 1}}}}{\text{ + b}}\dfrac{{{{\text{x}}^{{\text{1 + 1}}}}}}{{{\text{1 + 1}}}}{\text{ + cx + D}}\;$

${\text{ = }}\;\dfrac{{{\text{a}}{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ + }}\dfrac{{{\text{b}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + cx + D}}$

9. $\mathbf{\int {{\text{(2}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{e}}^{\text{x}}}{\text{)dx}}} }$

उत्तर: $\int {\left( {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{e}}^{\text{x}}}} \right)} {\text{dx}}\;$

\begin{align} \int {\left( {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{e}}^{\text{x}}}} \right)} {\text{dx}}\;{\text{ = }}\;{\text{2}}\int {{{\text{x}}^{\text{2}}}} {\text{dx + }}\int {{{\text{e}}^{\text{x}}}} {\text{dx}}\; \hfill \\ {\text{ = }}\;\dfrac{{{\text{2}} \times {{\text{x}}^{{\text{2 + 1}}}}}}{{{\text{2 + 1}}}}{\text{ + }}{{\text{e}}^{\text{x}}}{\text{ + C}}\;{\text{ = }}\;\dfrac{{{\text{2}}{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ + }}{{\text{e}}^{\text{x}}}{\text{ + C}} \hfill \\ \end{align}

10. $\mathbf{\int {{{{\text{(}}\sqrt {\text{x}} {\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}{\text{)}}}^{\text{2}}}{\text{dx}}} }$

उत्तर: $\int {{{\left( {\sqrt {\text{x}} {\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}} \right)}^{\text{2}}}} {\text{dx}}\;$

$\int {{{\left( {\sqrt {\text{x}} {\text{ - }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}} \right)}^{\text{2}}}} {\text{dx}}\;{\text{ = }}\;\int {\left[ {{{{\text{(}}\sqrt {\text{x}} {\text{)}}}^{\text{2}}}{\text{ - 2}}\sqrt {\text{x}} \times \dfrac{{\text{1}}}{{\sqrt {\text{x}} }}{\text{ + }}{{\left( {\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}} \right)}^{\text{2}}}} \right]} {\text{dx}}\;$

${\text{ = }}\;\int {\left( {{\text{x - 2 + }}\dfrac{{\text{1}}}{{\text{x}}}} \right)} {\text{dx}}\;{\text{ = }}\;\int {\text{x}} {\text{dx - 2}}\int {\text{1}} {\text{dx + }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {\text{dx}}\;$

${\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 2x + logx + C}}$

11. $\mathbf{\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 5}}{{\text{x}}^{\text{2}}}{\text{ - 4}}}}{{{{\text{x}}^{\text{2}}}}}{\text{dx}}} }$

उत्तर: $\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 5}}{{\text{x}}^{\text{2}}}{\text{ - 4}}}}{{{{\text{x}}^{\text{2}}}}}} {\text{dx}}\;$

$\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 5}}{{\text{x}}^{\text{2}}}{\text{ - 4}}}}{{{{\text{x}}^{\text{2}}}}}} {\text{dx}}\;{\text{ = }}\;\int {\dfrac{{{{\text{x}}^{\text{3}}}}}{{{{\text{x}}^{\text{2}}}}}} {\text{ + }}\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{4}}}{{{{\text{x}}^{\text{2}}}}}{\text{dx}}\;$

${\text{ = }}\;\int {\text{x}} {\text{dx + 5}}\int {\text{1}} {\text{dx - 4}}\int {{{\text{x}}^{{\text{ - 2}}}}} {\text{dx}}\;$

${\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + 5x - 4 }}\times \dfrac{{{{\text{x}}^{{\text{ - 2 + 1}}}}}}{{{\text{ - 2 + 1}}}}{\text{ + C}}\;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + 5x + }}\dfrac{{\text{4}}}{{\text{x}}}{\text{ + C}}$

12. $\mathbf{\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 3x + 4}}}}{{\sqrt {\text{x}} }}{\text{dx}}} }$

उत्तर: $\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 3x + 4}}}}{{\sqrt {\text{x}} }}} {\text{d}}\sqrt {\text{x}} \;$

\begin{align} \int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + 3x + 4}}}}{{\sqrt {\text{x}} }}} {\text{d}}\sqrt {\text{x}} \;{\text{ = }}\;\int {\left( {\dfrac{{{{\text{x}}^{\text{3}}}}}{{\sqrt {\text{x}} }}{\text{ + }}\dfrac{{{\text{3x}}}}{{\sqrt {\text{x}} }}{\text{ + }}\dfrac{{\text{4}}}{{\sqrt {\text{x}} }}} \right)} {\text{dx}}\;{\text{ = }}\;\int {\left( {{{\text{x}}^{{\text{3 - }}\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + }}} \right.} \left. {{\text{3}}{{\text{x}}^{{\text{1 - }}\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + 4}}{{\text{x}}^{{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}}}} \right){\text{dx}} \hfill \\ \;{\text{ = }}\;\int {{{\text{x}}^{\dfrac{{\text{5}}}{{\text{2}}}}}} {\text{dx + 3}}\int {{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}} {\text{dx + 4}}\int {{{\text{x}}^{\dfrac{{{\text{ - 1}}}}{{\text{2}}}}}} {\text{dx}} \hfill \\ \;{\text{ = }}\;\dfrac{{{{\text{x}}^{\dfrac{{\text{5}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{5}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + 3}}\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + 4}}\dfrac{{{{\text{x}}^{{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + C}} \hfill \\ \;{\text{ = }}\;\dfrac{{{{\text{x}}^{\dfrac{{\text{7}}}{{\text{2}}}}}}}{{\dfrac{{\text{7}}}{{\text{2}}}}}{\text{ + 3}}\dfrac{{{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + 4}}\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + C}} \hfill \\ {\text{ = }}\dfrac{{\text{2}}}{{\text{7}}}{\text{x7/2 + 2x3/2 + 8}}\sqrt {\text{x}} {\text{ + C}} \hfill \\ \end{align}

13. $\mathbf{\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ + x - 1}}}}{{{\text{x - 1}}}}{\text{dx}}} }$

उत्तर: $\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ + x - 1}}}}{{{\text{x - 1}}}}} {\text{dx}}$

\begin{align} \int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ + x - 1}}}}{{{\text{x - 1}}}}} {\text{dx}}\;{\text{ = }}\;\int {\dfrac{{{{\text{x}}^{\text{2}}}{\text{(x - 1) + 1(x - 1)}}}}{{{\text{(x - 1)}}}}} {\text{dx}} \hfill \\ \int {\dfrac{{{\text{(x - 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{{{\text{(x - 1)}}}}} {\text{dx}}\;{\text{ = }}\;\int {\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} {\text{dx = }}\int {{{\text{x}}^{\text{2}}}} {\text{dx + }}\int {\text{1}} {\text{dx}} \hfill \\ \;{\text{ = }}\;\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ + x + C}} \hfill \\ \end{align}

14. $\mathbf{\int {{\text{(1 - x)}}\sqrt {\text{x}} \;{\text{dx}}} }$

उत्तर: $\int {{\text{(1 - x)}}} \sqrt {\text{x}} {\text{dx}}$

\begin{align} \int {{\text{(1 - x)}}} \sqrt {\text{x}} {\text{dx}}\;{\text{ = }}\;\int {{\text{(}}\sqrt {\text{x}} {\text{ - x}}\sqrt {\text{x}} {\text{)}}} {\text{dx}} \hfill \\ {\text{ = }}\;\int {\left( {{{\text{x}}^{{\text{1/2}}}}{\text{ - }}{{\text{x}}^{{\text{1 + }}\dfrac{{\text{1}}}{{\text{2}}}}}} \right)} {\text{dx}} \hfill \\ {\text{ = }}\;\int {{{\text{x}}^{{\text{1/2}}}}} {\text{dx - }}\int {{{\text{x}}^{{\text{3/2}}}}} {\text{dx}} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ - }}\dfrac{{{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + C}} \hfill \\ {\text{ = }}\;\dfrac{{{{\text{x}}^{{\text{3/2}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ - }}\dfrac{{{{\text{x}}^{{\text{5/2}}}}}}{{\dfrac{{\text{5}}}{{\text{2}}}}}{\text{ + C}} \hfill \\ {\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}{{\text{x}}^{{\text{3/2}}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{5}}}{{\text{x}}^{{\text{5/2}}}}{\text{ + C}} \hfill \\ \end{align}

15. $\mathbf{\int {\sqrt {{\text{x(3}}{{\text{x}}^{\text{2}}}{\text{ + 2x + 3)}}} } }$

उत्तर: $\int {\sqrt {{\text{x}}\left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} \right){\text{dx}}} }$

\begin{align} \int {\sqrt {\text{x}} } \left( {{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} \right){\text{dx}}\;\; \hfill \\ {\text{ = }}\;\;\int {\left( {{\text{3}}{{\text{x}}^{{\text{2 + }}\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + 2}}{{\text{x}}^{{\text{1 + }}\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + 3}}{{\text{x}}^{{\text{1/2}}}}} \right)} {\text{dx}}\; \hfill \\ {\text{ = }}\;{\text{3}}\int {{{\text{x}}^{{\text{5/2}}}}} {\text{dx + 2}}\int {{{\text{x}}^{{\text{3/2}}}}} {\text{dx + 3}}\int {{{\text{x}}^{{\text{1/2}}}}} {\text{dx}}\; \hfill \\ {\text{ = }}\;{{3 \times }}\dfrac{{{{\text{x}}^{\dfrac{{\text{5}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{5}}}{{\text{2}}}{\text{ + 1}}}}{{ + 2 \times }}\dfrac{{{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}{\text{ + 1}}}}{{ \times }}\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + C}}\; \hfill \\ {\text{ = }}\;{\text{3}}\dfrac{{{{\text{x}}^{{\text{7/2}}}}}}{{\dfrac{{\text{7}}}{{\text{2}}}}}{\text{ + }}\dfrac{{{{\text{x}}^{{\text{5/2}}}}}}{{\dfrac{{\text{5}}}{{\text{2}}}}}{\text{ + 3}}\dfrac{{{{\text{x}}^{{\text{3/2}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + C}}\; \hfill \\ {\text{ = }}\;{{3 \times }}\dfrac{{\text{2}}}{{\text{7}}}{{\text{x}}^{{\text{7/2}}}}{{ + 2 \times }}\dfrac{{\text{2}}}{{\text{5}}}{{\text{x}}^{{\text{5/2}}}}{{ + 3 \times }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{x}}^{{\text{3/2}}}}{\text{ + C}}\; \hfill \\ {\text{ = }}\;\dfrac{{\text{6}}}{{\text{7}}}{{\text{x}}^{{\text{7/2}}}}{\text{ + }}\dfrac{{\text{4}}}{{\text{5}}}{{\text{x}}^{{\text{5/2}}}}{\text{ + 2}}{{\text{x}}^{{\text{3/2}}}}{\text{ + C}} \hfill \\ \end{align}

16. $\mathbf{\int {{\text{(2x - 3cos}}\;{\text{x + }}{{\text{e}}^{\text{3}}}{\text{)dx}}} }$

उत्तर: $\int {\left( {{\text{2x - 3cosx + }}{{\text{e}}^{\text{3}}}} \right)} {\text{dx}}\int {\left( {{\text{2x - 3cosx + }}{{\text{e}}^{\text{3}}}} \right)} {\text{dx}}$

\begin{align} {\text{ = }}\;{\text{2}}\int {\text{x}} {\text{dx - 3}}\int {{\text{cos}}} {\text{xdx + }}\int {{{\text{e}}^{\text{x}}}} {\text{dx2}}\dfrac{{{{\text{x}}^{\text{1}}}{\text{ + 1}}}}{{{\text{1 + 1}}}}{{ - 3 \times sinx + }}{{\text{e}}^{\text{x}}}{\text{ + C}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{2}}{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 3sinx + }}{{\text{e}}^{\text{x}}}{\text{ + C}}\; \hfill \\ {\text{ = }}\;\;{{\text{x}}^{\text{2}}}{\text{ - 3sinx + }}{{\text{e}}^{\text{x}}}{\text{ + C}} \hfill \\ \end{align}

17. $\mathbf{\int {{\text{(2}}{{\text{x}}^{\text{2}}}{\text{ - 3sin}}\;{\text{x + 5}}\sqrt {\text{x}} {\text{)dx}}} }$

उत्तर: $\int {\left( {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 3sinx + 5}}\sqrt {\text{x}} } \right)} {\text{dx}}$

\begin{align} \int {\left( {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 3sinx + 5}}\sqrt {\text{x}} } \right)} {\text{dx}}\;{\text{ = }}\;{\text{2}}\int {{{\text{x}}^{\text{2}}}} {\text{dx - 3}}\int {{\text{sin}}} {\text{xdx + 5}}\int {\sqrt {\text{x}} } {\text{dx}} \hfill \\ {\text{ = }}\;{{2 \times }}\dfrac{{{{\text{x}}^{{\text{2 + 1}}}}}}{{{\text{2 + 1}}}}{\text{ - 3( - cosx) + 5}}\int {{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}} {\text{dx}} \hfill \\ {\text{ = }}\;\dfrac{{{\text{2}}{{\text{x}}^{\text{3}}}}}{{\text{3}}}{{ + 3cosx + 5 \times }}\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + C}} \hfill \\ \end{align}

18. $\mathbf{\int {{\text{sec}}\;{\text{x}}\;{\text{(sec}}\;{\text{x}}\;{\text{ + }}\;{\text{tan}}\;{\text{x)dx}}}}$

उत्तर: $\int {{\text{sec}}} {\text{x}}\;{\text{(}}\;{\text{secx}}\;{\text{ + }}\;{\text{tanx}}\;{\text{)}}\;{\text{dx}}$

\begin{align} \left. {\int {{\text{sec}}} {\text{x}}\;{\text{(}}\;{\text{secx}}\;{\text{ + }}\;{\text{tanx}}\;{\text{)}}\;{\text{dx}}\;{\text{ = }}\;\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{x}}\;{\text{ + }}\;{\text{secx}}\;{\text{tanx}}} \right){\text{dx}} \hfill \\ {\text{ = }}\;\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{x}}\;{\text{dx}} \hfill \\ {\text{ = }}\;\int {{\text{sec}}} {\text{x}}\;{\text{tanx}}\;{\text{dx}} \hfill \\ {\text{ = }}\;{\text{tanx}}\;{\text{ + }}\;{\text{secx}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

19.$\mathbf{\int{\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{cose}}{{\text{c}}^{\text{2}}}{\text{x}}}}{\text{dx}}}}$

उत्तर: $\int {\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{cose}}{{\text{c}}^{\text{2}}}{\text{x}}}}} \;{\text{dx}}$

\begin{align} \int {\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{cose}}{{\text{c}}^{\text{2}}}{\text{x}}}}\;} {\text{dx}}\;{\text{ = }}\;\int {\dfrac{{\dfrac{{\text{1}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}}}{{\dfrac{{\text{1}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}}}}\;} {\text{dx}}\;{\text{ = }}\;\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \;{\text{dx}} \hfill \\ {\text{ = }}\;\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{x}}\;{\text{dx}} \hfill \\ \end{align}

\begin{align} {\text{ = }}\;\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - 1}}} \right)} \;{\text{dx}} \hfill \\ {\text{ = }}\;\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{x}}\;{\text{dx - }}\int {\text{1}} \;{\text{dx}} \hfill \\ {\text{ = }}\;{\text{tanx}}\;{\text{ - }}\;{\text{x}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

20. $\mathbf{\int {\dfrac{{{\text{2 - 3sin}}\;{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{dx}}} }$

उत्तर: $\int {\dfrac{{{\text{2 - 3sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \;{\text{dx}}\;$

\begin{align} \int {\dfrac{{{\text{2 - 3sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \;{\text{dx}}\;{\text{ = }}\;\int {\left( {\dfrac{{\text{2}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{ - }}\dfrac{{{\text{3sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \right)} \;{\text{dx}}\; \hfill \\ {\text{ = }}\;\int {\left( {\dfrac{{\text{2}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{\text{ - }}\dfrac{{{\text{3sinx}}}}{{{\text{cosx}}}}{{ \times }}\dfrac{{\text{1}}}{{{\text{cosx}}}}} \right)} \;{\text{dx}}\; \hfill \\ {\text{ = }}\;\int {\left( {{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x - 3secxtanx}}} \right)} \;{\text{dx}}\; \hfill \\ {\text{ = }}\;{\text{2}}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{x}}\;{\text{dx}}\;{\text{ - }}\;{\text{3}}\int {{\text{sec}}} {\text{xtanx}}\;{\text{dx}}\; \hfill \\ {\text{ = }}\;{\text{2tanx}}\;{\text{ - }}\;{\text{3secx}}\;{\text{ + C}} \hfill \\ \end{align}

### प्रशन 21 एवं 22 में सही उत्तर का चयन कीजिए-

21.  $\mathbf{{\text{(}}\sqrt {\text{x}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}{\text{)}}}$ का प्रति अवकलज है |

1. $\mathbf{\dfrac{{\text{1}}}{{\text{3}}}{{\text{x}}^{\dfrac{{\text{1}}}{{\text{3}}}}}{\text{ + 2}}{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + c}}}$

2. $\mathbf{\dfrac{{\text{3}}}{{\text{2}}}{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{x}}^{\text{2}}}{\text{ + c}}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\text{3}}}{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + 2}}{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + c}}}$

4. $\mathbf{\dfrac{{\text{3}}}{{\text{2}}}{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + c}}}$

उत्तर: $\left( {\sqrt {\text{x}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}} \right)\;{\text{ = }}\;\int {\left( {\sqrt {\text{x}} {\text{ + }}\dfrac{{\text{1}}}{{\sqrt {\text{x}} }}} \right)}$

\begin{align}{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{3/2}}}}{{\text{x}}^{{\text{3/2}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{1/2}}}}{{\text{x}}^{{\text{1/2}}}}{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}{{\text{x}}^{{\text{3/2}}}}{\text{ + 2}}{{\text{x}}^{{\text{1/2}}}}{\text{ + C}} \hfill \\ \end{align}

विकल्प (3) सही है |

22. यदि $\mathbf{\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{f(x)}}\;{\text{ = }}\;{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - }}\dfrac{{\text{3}}}{{{{\text{x}}^{\text{4}}}}}}$ जिसमे ${\text{f(2)}}\;{\text{ = }}\;{\text{0}}$ तो ${\text{f(x)}}$है |

1. $\mathbf{{{\text{x}}^{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ - }}\dfrac{{{\text{129}}}}{{\text{8}}}}$

2. $\mathbf{{{\text{x}}^{\text{3}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ + }}\dfrac{{{\text{129}}}}{{\text{8}}}}$

3. $\mathbf{{{\text{x}}^{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ + }}\dfrac{{{\text{129}}}}{{\text{8}}}}$

4. $\mathbf{{{\text{x}}^{\text{3}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ - }}\dfrac{{{\text{129}}}}{{\text{8}}}}$

उत्तर: $\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{f(x)}}\;{\text{ = }}\;{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - }}\dfrac{{\text{3}}}{{{{\text{x}}^{\text{4}}}}}$

\begin{align} {\text{ = }}\;{\text{f(x)}}\;{\text{ = }}\;\int {\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - }}\dfrac{{\text{3}}}{{{{\text{x}}^{\text{4}}}}}} \right)} \;{\text{dx}}\;{\text{ = }}\;\dfrac{{\text{4}}}{{\text{4}}}{{\text{x}}^{\text{4}}}{\text{ - }}\dfrac{{\text{3}}}{{{\text{ - 3}}}}{{ \times }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ + C}} \hfill \\{\text{ = }}\;{{\text{x}}^{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ + C}}\qquad \hfill \\ {\text{f(2)}}\;{\text{ = }}\;{\text{0}} \hfill \\{\text{ = }}\;{{\text{(2)}}^{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{{{{\text{(2)}}}^{\text{3}}}}}{\text{ + C = 0}}\quad \hfill \\ {\text{ = 16 + }}\dfrac{{\text{1}}}{{\text{8}}}{\text{ + C}}\;{\text{ = }}\;{\text{0 }} \hfill \\ \end{align}

\begin{align} {\text{5 - C}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{\text{129}}}}{{\text{8}}} \hfill \\{\text{ = }}\;{\text{f(x)}}\;{\text{ = }}\;{{\text{x}}^{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{3}}}}}{\text{ - }}\dfrac{{{\text{129}}}}{{\text{8}}} \hfill \\ \end{align}

विकल्प (1) सही है |

### प्रश्नावली 7.2

1 से 37 तक प्रश्नों में प्रत्येक फलन का समाकलन ज्ञात कीजिए :

1. $\mathbf{{\text{2x/1 + }}{{\text{x}}^{\text{2}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{(2}}} {\text{x/1 + }}{{\text{x}}^{\text{2}}}{\text{)dx}}$

मान लीजिए, ${\text{1 + }}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{2x}}\;{\text{dx = dt}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt}}\,{\text{/}}\;{\text{2x}} \hfill \\ \end{align}

\begin{align}\therefore \;{\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{2x}}}}{{{\text{t(2x)}}}}{\text{dt}}} \hfill \\{\text{ = }}\;\int {\text{1}} {\text{dt/t}}\;{\text{ = }}\;{\text{log|t|}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;{\text{log}}\left| {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right|\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

2. $\mathbf{{{\text{(logx)}}^{\text{2}}}{\text{/x}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{{{\text{(log}}\;{\text{x)}}}^{\text{2}}}}}{{\text{x}}}{\text{dx}}}$

मान लीजिए, ${\text{log}}\;{\text{x}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{1/x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{x}}\;{\text{dt}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{{{\text{t}}^{\text{2}}}{\text{x}}}}{{\text{x}}}{\text{dt}}} \; \hfill \\{\text{ = }}\;\int {{{\text{t}}^{\text{2}}}} {\text{dt}}\;{\text{ = }}\;{{\text{t}}^{\text{3}}}{\text{/3}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;{{\text{(log}}\;{\text{x)}}^{\text{3}}}{\text{/3}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

3.$\mathbf{{\text{1/(x + xlogx)}}}$

उत्तर: ${\text{I}}\,{\text{ = }}\;\int {{\text{(1}}} {\text{/x}}\;{\text{ + }}\;{\text{x}}\;{\text{log}}\;{\text{x)}}\;{\text{dx}}$

${\text{I}}\;{\text{ = }}\;\int {\text{1}} {\text{/x(1}}\;{\text{ + }}\;{\text{log}}\;{\text{x)}}\;{\text{dx}}$

मान लीजिए, ${\text{1 + log}}\;{\text{x}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{1/x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{xdt}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{1x}}}}{{\text{x}}}\dfrac{{{\text{dt}}}}{{\text{t}}}} \hfill \\{\text{ = }}\;\int {\text{1}} {\text{dt/t}}\;{\text{ = }}\;{\text{log|t|}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;{\text{log|1 + logx|}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

4. $\mathbf{{\text{sinx}}\;{\text{sin(cosx)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{(sin}}} \;{\text{x}}\;{\text{sin(cos}}\;{\text{x))dx}}$

मान लीजिए, ${\text{cos}}\;{\text{x}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{ - sin}}\;{\text{x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/ - sin}}\;{\text{x}} \hfill \\\therefore {\text{I = }}\int {\dfrac{{{\text{sin}}\;{\text{x}}{\text{.sin(t)}}}}{{{\text{ - sin}}\;{\text{x}}}}} {\text{dt}} \hfill \\{\text{ = }}\;{\text{ - }}\int {{\text{sin}}} {\text{t}}\;{\text{dt}}\;{\text{ = }}\;{\text{ - [ - cost]}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;{\text{cos(cosx)}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

5. $\mathbf{{\text{sin(ax + b)}}\;{\text{cos(ax + b)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{sin}}} {\text{(ax + b)}}\;{\text{cos(ax + b)}}\;{\text{dx}}$

\begin{align}{\text{sin(2A)}}\;{\text{ = }}\;{\text{2}}{\text{.sinA}}{\text{.cosA}} \hfill \\{\text{sin(ax + b)}}\;{\text{cos(ax + b)}}\;{\text{ = }}\;{\text{2}}{\text{.sin(ax + b)}}{\text{.cos(ax + b)/2}} \hfill \\{\text{ = }}\;{\text{sin2(ax + b)/2}} \hfill \\ \end{align}

मान लीजिए, ${\text{2(ax + b)}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{2a}}\;{\text{ = }}\;{\text{dt/dx}}\; \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/2a}} \hfill \\{\text{I}}\,{\text{ = }}\;\int {\dfrac{{{\text{sin}}\;{\text{t}}}}{{{\text{4a}}}}{\text{dt}}} \hfill \\{\text{ = }}\;\dfrac{{\text{1}}}{{{\text{4a}}}}\int {{\text{sint}}} \;{\text{dt}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\;{\text{1/4a[ - cos}}\;{\text{t]}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;{\text{ - (1/4a)}}\;{\text{cos2(ax + b)}}\;{\text{ + }}\;{\text{C }} \hfill \\ \end{align}

6. $\mathbf{\sqrt {{\text{ax}}} {\text{ + b}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\sqrt {{\text{ax + b}}} } \;{\text{dx}}$

\begin{align}{\text{ = }}\;\int {{{{\text{(ax + b)}}}^{{\text{1/2}}}}} {\text{dx}} \hfill \\{\text{ = }}\;{{a \times }}\dfrac{{{{{\text{(ax + b)}}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}\;{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}{\text{a(ax + b}}{{\text{)}}^{{\text{3/2}}}}{\text{ + c}} \hfill \\ \end{align}

7. $\mathbf{{\text{x}}\sqrt {\text{x}} {\text{ + 2}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\text{x}} \sqrt {\text{x}} {\text{ + 2}}\;{\text{dx}}$

मान लीजिए, ${\text{x}}\;{\text{ + }}\;{\text{2}}\;{\text{ = }}\;{\text{t}}$ तथा ${\text{x}}\;{\text{ = }}\;{\text{t}}\;{\text{ - }}\;{\text{2}}$

\begin{align}{\text{1}}\;{\text{ = }}\;{\text{dt}}\;{\text{/}}\;{\text{dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt}} \hfill \\{\text{ = }}\;\int {{\text{(t - 2)}}} \sqrt {\text{t}} \;{\text{dt}} \hfill \\{\text{ = }}\;\int {{\text{(t - 2)}}} {{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}\;{\text{dt}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\;\left( {{{\text{t}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\;{\text{ - }}\;{\text{2}}{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}} \right){\text{dt}} \hfill \\{\text{ = }}\;\dfrac{{{{\text{t}}^{{\text{1 +}}\dfrac{{\text{3}}}{{\text{2}}}}}}}{{\dfrac{{\text{3}}}{{\text{2}}}{\text{ + 1}}}}\;{\text{ - }}\;\dfrac{{{\text{2}}{{\text{t}}^{{\text{1 + }}\dfrac{{\text{1}}}{{\text{2}}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}\;{\text{ + }}\;{\text{c}} \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{5}}}{{\text{t}}^{\dfrac{{\text{5}}}{{\text{2}}}}}\;{\text{ - }}\;\dfrac{{\text{4}}}{{\text{3}}}{{\text{t}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\;{\text{ + }}\;{\text{C}} \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{5}}}{{\text{(x + 2)}}^{\dfrac{{\text{5}}}{{\text{2}}}}}\;{\text{ - }}\;\dfrac{{\text{4}}}{{\text{3}}}{{\text{(x + 2)}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

8. $\mathbf{{\text{x}}\sqrt {{\text{1 + 2}}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{x}}\sqrt {{\text{1 + 2}}{{\text{x}}^{\text{2}}}} } {\text{dx}}$

मान लीजिए, ${\text{1 + 2}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{4x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/4x}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{x}}\sqrt {\text{t}} }}{{\text{4}}}\dfrac{{{\text{dt}}}}{{\text{x}}}} \hfill \\{\text{ = }}\;{\text{1/4}}\int {\sqrt {\text{t}} } \;{\text{dt}} \hfill \\{\text{ = }}\;{\text{1/4}}\int {{{\text{t}}^{{\text{1/2}}}}} {\text{dt}} \hfill \\{\text{ = }}\;{\text{1}}{{\text{t}}^{{\text{1 + 1/2}}}}{\text{/4(1/2) + 1 + C}} \hfill \\{\text{ = }}\;{\text{1/6}}{\left( {{\text{1 + 2}}{{\text{x}}^{\text{2}}}} \right)^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + C}} \hfill \\ \end{align}

9. $\mathbf{{\text{(4x + 2)}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} }$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{(4x + 2)}}} \sqrt {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \;{\text{dx}}$

मान लीजिए, ${{\text{x}}^{\text{2}}}{\text{ + x + 1}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{2x + 1 = }}\dfrac{{{\text{dt}}}}{{{\text{dx}}}} \hfill \\{\text{dx}}\;{\text{ = }}\;\dfrac{{{\text{dt}}}}{{{\text{(2x + 1)}}}} \hfill \\\therefore {\text{I = }}\;\int {\dfrac{{{\text{(4x + 2)}}\sqrt {\text{t}} }}{{{\text{(2x + 1)}}}}{\text{dt}}} \hfill \\{\text{ = }}\;\dfrac{{{\text{2(2x + 1)}}\sqrt {\text{t}} }}{{{\text{(2x + 1)}}}}{\text{dt}} \hfill \\{\text{ = }}\;{\text{2}}\int {\sqrt {\text{t}} } {\text{dt}} \hfill \\{\text{ = }}\;{{2 \times }}\dfrac{{{{\text{t}}^{{\text{1 + }}\dfrac{{\text{1}}}{{\text{2}}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{\text{4}}}{{\text{3}}}{\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right)^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + C}} \hfill \\ \end{align}

10. $\mathbf{{\text{1/(x - }}\sqrt {\text{x}} {\text{)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\text{1}} {\text{/x - }}\sqrt {\text{x}} \;{\text{dx}}$

${\text{ = }}\;\int {\text{1}} {\text{/}}\sqrt {\text{x}} {\text{(}}\sqrt {\text{x}} {\text{ - 1)dx}}$

मान लीजिए, $\sqrt {\text{x}} {\text{ - 1}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{1/2}}\sqrt {\text{x}} \;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{2}}\sqrt {\text{x}} {\text{dt}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{1}}}{{\sqrt {{\text{xt}}} {\text{ 2}}\sqrt {\text{x}} {\text{ }}}}} {\text{ dt}} \hfill \\{\text{ = }}\;\int {\text{2}} \dfrac{{{\text{dt}}}}{{\text{t}}}\;{\text{ = }}\;{{2 \times log|t|}}\;{\text{ + }}\;{\text{C}}\; \hfill \\{\text{ = }}\;{\text{2log|}}\sqrt {{\text{x - 1}}} {\text{|}}\;{\text{ + }}\;{\text{C }} \hfill \\ \end{align}

11. $\mathbf{{\text{x/}}\sqrt {\text{x}} {\text{ + 4}}\;{\text{,}}\;{\text{x > 0}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{\sqrt {\text{x}} }}} \;{\text{ + }}\;{\text{4}}\;{\text{dx}}$

मान लीजिए, ${\text{x + 4}}\;{\text{ = }}\;{\text{t}}$ तथा ${\text{x}}\;{\text{ = }}\;{\text{t - 4}}$

\begin{align}{\text{dx}}\;{\text{ = }}\;{\text{dt}} \hfill \\{\text{I}}\;{\text{ = }}\;\int {\text{t}} {\text{ - }}\dfrac{{\text{4}}}{{\sqrt {\text{t}} }}{\text{dt}} \hfill \\{\text{ = }}\;\int {\sqrt {\text{t}} } {\text{ - }}\dfrac{{\text{4}}}{{\sqrt {\text{t}} }}{\text{dt}} \hfill \\{\text{ = }}\;\int {{\text{(}}{{\text{t}}^{{\text{1/2}}}}} \;{\text{ - }}\;{\text{4}}{{\text{t}}^{\dfrac{{{\text{ - 1}}}}{{\text{2}}}}}{\text{)dt}} \hfill \\{\text{ = }}\;\dfrac{{{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}}}}}{{\dfrac{{\text{1}}}{{\text{2}}}}}\;{\text{ + }}\;{\text{1}}\;{\text{ - }}\;{\text{4}}\dfrac{{{{\text{t}}^{\dfrac{{{\text{ - 1}}}}{{\text{2}}}{\text{ + 1}}}}}}{{{\text{(}}\dfrac{{{\text{ - 1}}}}{{\text{2}}}{\text{ + 1)}}}}\;{\text{ + }}\;{\text{C}} \hfill \\\hfill \\ \end{align}

\begin{align}{\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}{{\text{t}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ - 8}}{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + C}} \hfill \\{\text{ = }}\;{\text{t}}{\text{.}}{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}\dfrac{{\text{2}}}{{\text{3}}}{\text{ - 8}}{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{(t - 12)}}\;{\text{ + }}\,{\text{C}} \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}\sqrt {{\text{x + 4}}} {\text{(x - 8)}}\;{\text{ + }}\;{\text{C}} \hfill \\ \end{align}

12. $\mathbf{{\left( {{{\text{x}}^{\text{3}}}{\text{ - 1}}} \right)^{{\text{1/3}}}}{{\text{x}}^{\text{5}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{{\left( {{{\text{x}}^{\text{3}}}{\text{ - 1}}} \right)}^{{\text{1/3}}}}} {{\text{x}}^{\text{5}}}{\text{dx}}$

${\text{ = }}\;{\left( {{{\text{x}}^{\text{3}}}{\text{ - 1}}} \right)^{{\text{1/3}}}}{{\text{x}}^{\text{3}}}{\text{ \times }}{{\text{x}}^{\text{2}}}{\text{dx}}$

मान लीजिए, ${{\text{x}}^{\text{3}}}{\text{ - 1}}\;{\text{ = }}\;{\text{t}}$ तथा ${{\text{x}}^{\text{3}}}\;{\text{ = }}\;{\text{t + 1}}$

\begin{align}{\text{3}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/3}}{{\text{x}}^{\text{2}}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{{t^{\dfrac{1}{3}}}(t + 1){x^2}}}{{3{x^2}}}\;dt} \hfill \\{\text{ = }}\;\dfrac{1}{3}\int {\left( {{{\text{t}}^{\dfrac{4}{3}}}{\text{ + }}{{\text{t}}^{\dfrac{1}{3}}}} \right)} {\text{dt}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\;\dfrac{1}{3}\left[ {\dfrac{{{{\text{t}}^{\dfrac{7}{3}}}}}{{\dfrac{7}{3}}} + \dfrac{{{t^{\dfrac{4}{3}}}}}{{\dfrac{4}{3}}}} \right] + C \hfill \\{\text{ = }}\;\dfrac{1}{3}\left[ {\dfrac{3}{7}{{\text{t}}^{\dfrac{7}{3}}}{\text{ + }}\dfrac{3}{4}{{\text{t}}^{\dfrac{4}{3}}}} \right]{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{1}{7}{{\text{t}}^{\dfrac{7}{3}}}{\text{ + }}\dfrac{1}{4}{{\text{t}}^{\dfrac{4}{3}}}{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{1}{7}{\left( {{{\text{x}}^{\text{3}}}{\text{ - 1}}} \right)^{\dfrac{7}{3}}}{\text{ + }}\dfrac{1}{4}{\left( {{{\text{x}}^{\text{3}}}{\text{ - 1}}} \right)^{\dfrac{4}{3}}}{\text{ + C}} \hfill \\ \end{align}

13. $\mathbf{{{\text{x}}^{\text{2}}}{\text{/}}{\left( {{\text{2 + 3}}{{\text{x}}^{\text{3}}}} \right)^{\text{3}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{{\text{x}}^{\text{2}}}} {\text{/}}{\left( {{\text{2 + 3}}{{\text{x}}^{\text{3}}}} \right)^{\text{3}}}{\text{dx}}$

मान लीजिए, ${\text{2 + 3}}{{\text{x}}^{\text{3}}}{\text{ = }}\;{\text{t}}$

\begin{align}{\text{9}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/9}}{{\text{x}}^{\text{2}}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {{{\text{x}}^{\text{2}}}} {\text{dt/}}{{\text{t}}^{\text{3}}}{\text{9}}{{\text{x}}^{\text{2}}}{\text{ }} \hfill \\{\text{ = }}\;{\text{1/9}}\int {\text{d}} {\text{t/}}{{\text{t}}^{\text{3}}}\;{\text{ = }}\;{\text{1/9}}\int {{{\text{t}}^{{\text{ - 3}}}}} {\text{dt}} \hfill \\{\text{ = }}\;{\text{1/9}}\left( {\dfrac{{{{\text{t}}^{{\text{ - 3 + 1}}}}}}{{ - 3 + 1}}} \right){\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{{{\text{t}}^{{\text{ - 2}}}}}}{9}{{ \times ( - 2) + C}} \hfill \\{\text{ = }}\;\dfrac{{ - 1}}{{18}}{{\text{t}}^{\text{2}}}{\text{ + C}} \hfill \\{\text{ = }}\;{\text{ - }}\dfrac{1}{{18}}{\left( {{\text{2 + 3}}{{\text{x}}^{\text{3}}}} \right)^{\text{2}}}{\text{ + }}\,{\text{C}} \hfill \\ \end{align}

14. $\mathbf{{\text{1/x(logx}}{{\text{)}}^{\text{m}}}\;{\text{,}}\;{\text{x > 0}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\text{1}} {\text{/x(logx}}{{\text{)}}^{\text{m}}}{\text{dx}}$

मान लीजिए, ${\text{log}}\;{\text{x}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{1/x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{xdt}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\text{x}} {\text{dt/x}}{{\text{t}}^{\text{m}}} \hfill \\{\text{ = }}\;\int {{{\text{t}}^{{\text{ - m}}}}} {\text{dt}}\;{\text{ = }}\;{{\text{t}}^{{\text{ - m + 1}}}}{\text{/ - m + 1 + C}} \hfill \\{\text{ = }}\;\dfrac{{{{{\text{(log}}\;{\text{x)}}}^{{\text{1 - m}}}}}}{{1 - m}}{\text{ + C}} \hfill \\ \end{align}

15. $\mathbf{{\text{x/9 - 4}}{{\text{x}}^{\text{2}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{{\text{9 - 4}}{{\text{x}}^{\text{2}}}}}\;} {\text{dx}}$

मान लीजिए, ${\text{9 - 4}}{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{ - 8x}}\;{\text{ = }}\;\dfrac{{{\text{dt}}}}{{{\text{dx}}}} \hfill \\{\text{dx = }}\;\dfrac{{{\text{dt}}}}{{{\text{ - 8}}}} \hfill \\ \end{align}

\begin{align}\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{{\text{ - 8x}}}}\dfrac{{{\text{dt}}}}{{\text{t}}}} \hfill \\{\text{ = }}\;{\text{ - 1/8}}\int {{\text{dt}}} {\text{/t}} \hfill \\{\text{ = }}\;\dfrac{{{\text{log|t|}}\;}}{{{\text{ - 8}}}}\;{\text{ + }}\,{\text{C}} \hfill \\{\text{ = }}\;\dfrac{{{\text{ - log}}\left| {{\text{9 - 4}}{{\text{x}}^{\text{2}}}} \right|}}{{\text{8}}}{\text{ + C}} \hfill \\ \end{align}

16. $\mathbf{{{\text{e}}^{{\text{2x + 3}}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{{\text{e}}^{{\text{2x + 3}}}}} {\text{dx}}$

मान लीजिए, ${\text{2x + 3}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{2}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/2}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{{{\text{e}}^{\text{t}}}}}{2}} {\text{dt}} \hfill \\{\text{ = }}\;{\text{1/2}}\int {{{\text{e}}^{\text{t}}}} {\text{dt = }}\;\dfrac{{{{\text{e}}^{\text{t}}}}}{2}{\text{ + C}} \hfill \\{\text{ = }}\;{{\text{e}}^{{\text{2x + 3}}}}{\text{/2 + C}} \hfill \\ \end{align}

17. $\mathbf{{\text{x/}}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\text{x}} {\text{/}}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{dx}}$

मान लीजिए, ${{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{t}}$

\begin{align} {\text{2x}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;{\text{dt/2x }} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{{{\text{e}}^{\text{t}}}{\text{2x}}}}{\text{dt}}} \hfill \\{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\int {{{\text{e}}^{{\text{ - t}}}}} {\text{dt}} \hfill \\{\text{ = }}\;\dfrac{{{\text{ - }}{{\text{e}}^{{\text{ - t}}}}}}{{\text{2}}}{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{{\text{ - }}{{\text{e}}^{{\text{ - }}{{\text{x}}^{\text{2}}}}}}}{{\text{2}}}{\text{ + C}} \hfill \\{\text{ = }}\;\dfrac{{{\text{ - 1}}}}{{{\text{2}}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}}}{\text{ + C}} \hfill \\ \end{align}

18. $\mathbf{{{\text{e}}^{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{\text{/1 + }}{{\text{x}}^{\text{2}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\,\int {\dfrac{{{{\text{e}}^{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}\,} {\text{dx}}$

मान लीजिए, ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\,{\text{ = }}\,{\text{t}}$

\begin{align}{\text{1/1 + }}{{\text{x}}^{\text{2}}}\,{\text{ = }}\,{\text{dt/dx}} \hfill \\{\text{dx}}\,{\text{ = }}\,\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right){\text{dt}} \hfill \\\therefore {\text{I}}\,{\text{ = }}\,\int {\dfrac{{{{\text{e}}^{\text{t}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right){\text{dt}} \hfill \\{\text{ = }}\,\int {{{\text{e}}^{\text{t}}}} {\text{dt = }}\,{{\text{e}}^{\text{t}}}{\text{ + C}} \hfill \\{\text{ = }}\,{{\text{e}}^{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{\text{ + C}} \hfill \\ \end{align}

19. $\mathbf{{{\text{e}}^{{\text{2x}}}}{\text{ - 1/}}{{\text{e}}^{{\text{2x}}}}{\text{ + 1}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\left( {{{\text{e}}^{{\text{2x}}}}{\text{ - 1}}} \right)} {\text{/}}{{\text{e}}^{{\text{2x + 1}}}}{\text{dx}}$

\begin{align}{\text{ = }}\;\int {\left( {{{\text{e}}^{\text{x}}}{{ \times }}{{\text{e}}^{\text{x}}}{\text{ - 1}}} \right)} {\text{/}}\left( {{{\text{e}}^{\text{x}}}{{ \times }}{{\text{e}}^{\text{x}}}{\text{ + 1}}} \right){\text{dx}} \hfill \\{\text{ = }}\;\int {{{\text{e}}^{\text{x}}}} \left( {{{\text{e}}^{\text{x}}}{\text{ - 1/}}{{\text{e}}^{\text{x}}}} \right){\text{/}}{{\text{e}}^{\text{x}}}\left( {{{\text{e}}^{\text{x}}}{\text{ + 1/}}{{\text{e}}^{\text{x}}}} \right){\text{dx}} \hfill \\{\text{ = }}\;\int {\left( {{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}} \right)} {\text{/}}\left( {{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}} \right){\text{dx}} \hfill \\ \end{align}

मान लीजिए, ${{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{{\text{e}}^{{\text{ex}}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}\;{\text{ = }}\;{\text{dt/dx}} \hfill \\{\text{dx}}\;{\text{ = }}\;\dfrac{{{\text{dt}}}}{{{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}}} \hfill \\\therefore {\text{I}}\;{\text{ = }}\;\int {\dfrac{{\left( {{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}} \right)}}{{\left( {{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}} \right)}}} \dfrac{{{\text{dt}}}}{{\text{t}}} \hfill \\{\text{ = }}\;\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} \;{\text{ = }}\;{\text{log|t| + C}} \hfill \\{\text{ = }}\;{\text{log}}\left| {{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}} \right|{\text{ + C}} \hfill \\ \end{align}

20. $\mathbf{{{\text{e}}^{{\text{2x}}}}{\text{ - }}{{\text{e}}^{{\text{ - 2x}}}}{\text{/}}{{\text{e}}^{{\text{2x}}}}{\text{ + }}{{\text{e}}^{{\text{ - 2x}}}}}$

उत्तर: ${\text{I}}\,{\text{ = }}\,\int {\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{ - }}{{\text{e}}^{{\text{ - 2x}}}}}}{{{{\text{e}}^{{\text{2x}}}}{\text{ + }}{{\text{e}}^{{\text{ - 2x}}}}}}} {\text{dx}}$

मान लीजिए, ${{\text{e}}^{{\text{2x}}}}{\text{ + }}{{\text{e}}^{{\text{ - 2x}}}}\;{\text{ = }}\,{\text{t}}$

\begin{align}{\text{2}}{{\text{e}}^{{\text{2x}}}}{\text{ + 2}}{{\text{e}}^{{\text{ - 2x}}}}\,{\text{ = }}\,{\text{dt/dx}} \hfill \\{\text{dx}}\,{\text{ = }}\,{\text{dt/2}}\left( {{{\text{e}}^{{\text{2x}}}}{\text{ - }}{{\text{e}}^{{\text{ - 2x}}}}} \right) \hfill \\\therefore {\text{I}}\,{\text{ = }}\,\int {\dfrac{{{\text{(}}{{\text{e}}^{{\text{2x}}}}{\text{ - }}{{\text{e}}^{{\text{ - 2x}}}}{\text{)}}}}{{{\text{2}}\left( {{{\text{e}}^{{\text{2x}}}}{\text{ - }}{{\text{e}}^{{\text{ - 2x}}}}} \right)}}} \dfrac{{{\text{dt}}}}{{\text{t}}} \hfill \\{\text{ = }}\,\dfrac{{\text{1}}}{{\text{2}}}\int {\text{d}} {\text{t/t}} \hfill \\{\text{ = }}\,\dfrac{{{\text{log|t|}}}}{{\text{2}}}{\text{ + C}} \hfill \\ \end{align}

${\text{ = }}\,\dfrac{{{\text{log|}}{{\text{e}}^{{\text{2x}}}}{\text{ + }}{{\text{e}}^{{\text{ - 2x}}}}{\text{|}}}}{{\text{2}}}{\text{ + C}}$

21. $\mathbf{{\text{ta}}{{\text{n}}^{\text{2}}}{\text{(2x - 3)}}}$

उत्तर: ${\text{I = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(2x - 3)dx}}$

\begin{align}{\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{(2x - 3)dx - }}\int {\text{1}} {\text{dx}} \hfill \\ {\text{2x - 3 = t}}\qquad \hfill \\\begin{array}{*{20}{c}} {{\text{2dx = dt}}} \\ {{\text{dx = dt/2}}} \end{array} \hfill \\\therefore {\text{I}}\;{\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{t dt/2 - }}\int {\text{1}} {\text{ dx}} \hfill \\{\text{ = 1/2}}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{t dt - }}\int {\text{1}} {\text{ dx }} \hfill \\{\text{ = (1/2)(tant - x + C)}} \hfill \\{\text{ = (1/2)[tan(2x - 3) - x + C}} \hfill \\ \end{align}

22. $\mathbf{{\text{se}}{{\text{c}}^{\text{2}}}{\text{(7 - 4x)}}}$

उत्तर: ${\text{I = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{(7 - 4x) dx}}$

\begin{align}{\text{7 - 4x = t}} \hfill \\{\text{ - 4 = dt/dx}} \hfill \\{\text{dx = dt/ - 4}} \hfill \\{\text{l = }}\dfrac{{ - 1}}{4}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{t dt}} \hfill \\{\text{ = - }}\dfrac{{{\text{tan}}\;{\text{t}}}}{4}{\text{ + C}} \hfill \\{\text{ = - }}\dfrac{1}{4}{\text{tan(7 - 4x) + C}} \hfill \\ \end{align}

23. $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x/}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$

माना      ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\;{\text{ = }}\;{\text{t}}$

$\begin{array}{*{20}{r}}{{\text{1/}}\sqrt {\text{1}} {\text{ - }}{{\text{x}}^{\text{2}}}{\text{ = dt/dx}}} \\ {{\text{dx = }}\sqrt {\text{1}} {\text{ - }}{{\text{x}}^{\text{2}}}{\text{dt}}} \\ {\therefore \;{\text{I = }}\int {\text{t}} \sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{dt/}}\sqrt {\text{1}} {\text{ - }}{{\text{x}}^{\text{2}}}} \\ {{\text{ = }}\int {\text{t}} {\text{ dt = }}{{\text{t}}^{\text{2}}}{\text{/2 + C}}} \\ {{\text{ = }}{{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)}^{\text{2}}}{\text{/2 + C}}} \end{array}$

24. $\mathbf{{\text{2cosx - 3sinx/6cosx + 4sinx}}}$

उत्तर: ${\text{I = }}\int {{\text{[(2cosx - 3sinx)}}} {\text{/(6cosx + 4sinx)]dx}}$

\begin{align}{\text{ = }}\int {{\text{[(2cosx - 3sinx)}}} {\text{/2(3cosx + 2sinx)]dx}} \hfill \\{\text{3cosx + 2sinx = t}} \hfill \\{\text{ - 3sinx + 2cosx = dt/dx}} \hfill \\{\text{dx = dt/(2cosx - 3sinx)}} \hfill \\ \end{align}

\begin{align}\therefore \;{\text{I = }}\int {\dfrac{{{\text{(2cosx - 3sinx)}}}}{{{\text{2t(2cosx - 3sinx)}}}}\;{\text{dt}}} \hfill \\ {\text{ = }}\dfrac{1}{2}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} \hfill \\ \end{align}

${\text{ = }}\dfrac{1}{2}{\text{log|t| + C}}$

${\text{ = }}\dfrac{1}{2}{\text{log|3cosx + 2sinx| + C}}$

25. $\mathbf{{\text{1/co}}{{\text{s}}^{\text{2}}}{\text{x(1 - tanx}}{{\text{)}}^{\text{2}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{1}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x(1 - tanx}}{{\text{)}}^{\text{2}}}}}{\text{dx}}\;{\text{ = }}\;\int {\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{{{\text{(1 - tanx)}}}^{\text{2}}}}}} {\text{dx}}}$

माना               ${\text{1 - tanx}}\;{\text{ = }}\;{\text{t}}$

\begin{array}{*{20}{r}}{{\text{ - se}}{{\text{c}}^{\text{2}}}{\text{x = dt/dx}}} \\ {{\text{ dx = dt/ - se}}{{\text{c}}^{\text{2}}}{\text{x}}} \\ {\therefore {\text{ I = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{xdt/}}{{\text{t}}^{\text{2}}}\left( {{\text{ - se}}{{\text{c}}^{\text{2}}}{\text{x}}} \right)} \\ \begin{align}{\text{ = - }}\int {{{\text{t}}^{{\text{ - 2}}}}} {\text{dt}} \hfill \\{\text{ = - }}\left( {{\text{ - }}{{\text{t}}^{{\text{ - 2 + 1}}}}{\text{/ - 2 + 1}}} \right){\text{ + C}} \hfill \\ \end{align} \end{array}

26. $\mathbf{{\text{cos}}\sqrt {\text{x}} {\text{/}}\sqrt {\text{x}} }$

उत्तर: ${\text{I}}\;{\text{ = }}\;\dfrac{{{\text{cos}}\sqrt {\text{x}} }}{{\sqrt {\text{x}} }}$

माना  $\sqrt {\text{x}} \; = \;{\text{t}}$

\begin{align}{\text{1/2}}\sqrt {\text{x}} {\text{ = dt/dx}} \hfill \\{\text{dx = 2}}\sqrt {\text{x}} {\text{dt}} \hfill \\\therefore \;{\text{I = }}\int {\dfrac{{\cos 2\sqrt x }}{{\sqrt x }}} {\text{dt}} \hfill \\{\text{ = 2}}\int {{\text{cos}}} {\text{tdt}} \hfill \\{\text{ = 2sint + C}} \hfill \\{\text{ = 2sin}}\sqrt {\text{x}} {\text{ + C}} \hfill \\ \end{align}

27. $\mathbf{\sqrt {{{sin2x \times cos2x}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{sin2x}}} } {\text{ cos2x dx}}$

माना ${\text{sin2x = t}}$

\begin{align}{\text{2cos2x = dt/dx}} \hfill \\{\text{dx = dt/(2cos2x)}} \hfill \\ \therefore \;{\text{I = }}\dfrac{1}{2}\int {\sqrt {\text{t}} } {\text{(cos2x/cos2x)dt}} \hfill \\\end{align}

\begin{align}{\text{ = }}\dfrac{1}{2}\int {\sqrt {\text{t}} } {\text{ dt}} \hfill \\{\text{ = }}{{\text{t}}^{{\text{1/2}}}}{\text{ + }}\dfrac{1}{2}{\text{(}}\dfrac{1}{2}{\text{ + 2) + C}} \hfill \\ \end{align}

${\text{ = }}{{\text{t}}^{3/2}}{\text{ + }}\dfrac{5}{4}{\text{ + C}}$

${\text{ = }}\dfrac{1}{3}{{\text{(sin2x)}}^{3/2}}{\text{ + C}}$

28. $\mathbf{{\text{cosx/}}\sqrt {{\text{1 + sin}}\;{\text{x}}}}$

उत्तर: ${\text{I = }}\int {{\text{cos}}} {\text{x/}}\sqrt {1 + \sin {\text{x}}} {\text{ dx}}$ ${\text{ = }}\int {{{{\text{(1 + sinx)}}}^{{\text{ - 1/2}}}}} {\text{cosx dx}}$

माना ${\text{1 + sinx = t}}$

\begin{align}{\text{cosx = dt/dx}} \hfill \\{\text{dx = dt/cosx}} \hfill \\\therefore {\text{ I = }}\int {{{\text{t}}^{{\text{ - 1/2}}}}{\text{cosx}}\dfrac{{{\text{dt}}}}{{{\text{cosx}}}}} \hfill \\{\text{ = }}\dfrac{{{{\text{t}}^{{\text{( - 1/2) + 1}}}}}}{{{\text{( - 1/2) + 1}}}}{\text{ + C}} \hfill \\{\text{ = 2}}{{\text{t}}^{1/2}}{\text{ + C}} \hfill \\{\text{ = 2}}\sqrt {{\text{1 + sinx}}} {\text{ + C}} \hfill \\ \end{align}

29. $\mathbf{{\text{cotx}}\;{\text{log(sinx)}}}$

उत्तर: ${\text{I = }}\int {{\text{cot}}} {\text{x log(sinx)dx}}$

माना ${\text{log(sinx) = t}}$

\begin{align}{\text{cosx/sinx = dt/dx}} \hfill \\{\text{cotx = dt/dx}} \hfill \\{\text{dx = dt/cotx}} \hfill \\\therefore {\text{ I = }}\int {{\text{(cot}}} {\text{x)t}}\dfrac{{{\text{dt}}}}{{{\text{cotx}}}} \hfill \\{\text{ = }}\int {\text{t}} {\text{ dt}} \hfill \\{\text{ = }}{{\text{t}}^{\text{2}}}{\text{/2 + C}} \hfill \\{\text{ = (log(sinx)}}{{\text{)}}^{\text{2}}}{\text{/2 + C}} \hfill \\ \end{align}

30. $\mathbf{{\text{sin}}\;{\text{x/(1 + cos}}\;{\text{x)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{sinx}}}}{{{\text{1 + cosx}}}}{\text{dx}}}$

माना ${\text{1 + cosx = t}}$

\begin{align}{\text{ - sinx = dt/dx}} \hfill \\{\text{dx = dt/( - sinx)}} \hfill \\\therefore {\text{ I = }}\int {\dfrac{{{\text{sinx}}}}{{\text{t}}}\dfrac{{{\text{dt}}}}{{{\text{( - sinx)}}}}} \hfill \\ \end{align}

${\text{ = - }}\int {\text{d}} {\text{t/t}}$

$= \; - \log \left| {\text{t}} \right| + {\text{C}}$

${\text{ = }}\;{\text{ - log}}\left| {{\text{1 + cosx}}} \right|{\text{ + C}}$

31. $\mathbf{{\text{sin}}\;{\text{x}}\;{\text{/}}\;{{\text{(1 + cos}}\;{\text{x)}}^{\text{2}}}}$

उत्तर: ${\text{I = }}\int {{\text{sin}}} {\text{x/(1 + cosx}}{{\text{)}}^{\text{2}}}{\text{ dx}}$

माना ${\text{1 + cosx = t}}$

\begin{align}{\text{ - sinx = dt/dx}} \hfill \\{\text{dx = dt/( - sinx)}} \hfill \\\therefore \;{\text{I = }}\int {{\text{(sin}}} {\text{xdt)/[}}{{\text{t}}^{\text{2}}}{\text{( - sinx)]}} \hfill \\{\text{ = - }}\int {\text{d}} {\text{t/}}{{\text{t}}^{\text{2}}} \hfill \\{\text{ = - }}\int {{{\text{t}}^{{\text{ - 2}}}}} {\text{dt}} \hfill \\{\text{ = - }}{{\text{t}}^{{\text{ - 2 + 1}}}}{\text{/( - 2 + 1) + C}} \hfill \\{\text{ = (1/t) + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{1 + cosx}}}}{\text{ + C}} \hfill \\ \end{align}

32. $\mathbf{{\text{1}}\;{\text{/}}\;({\text{1}}\;{\text{ + }}\;{\text{cot}}\;{\text{x)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{\text{[1/(1 + cosx)]dx}}}$

\begin{align}{\text{ = }}\int {\text{1}} {\text{/(1 + cosx/sinx)dx}} \hfill \\{\text{ = }}\int {{\text{sin}}} {\text{x/(sinx + cosx)dx}} \hfill \\{\text{ = }}\int {\text{2}} {\text{sinx/[2(sinx + cosx)]dx}} \hfill \\{\text{ = }}\dfrac{1}{2}\int {\text{2}} {\text{sinx/(sinx + cosx)dx}} \hfill \\{\text{ = }}\dfrac{1}{2}\int {{\text{(sinx + sinx + cosx - cosx)}}} {\text{/(sinx + cosx)}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{[}}\int {\dfrac{{{\text{sinx + cosx}}}}{{{\text{sinx + cosx}}}}{\text{dx + }}\int {\dfrac{{{\text{sinx - cosx}}}}{{{\text{sinx + cosx}}}}{\text{dx]}}} } \hfill \\{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{[}}\int {{\text{dx}}} {\text{ + }}\int {\dfrac{{{\text{sinx - cosx}}}}{{{\text{sinx + cosx}}}}{\text{dx]}}} \hfill \\ \end{align}

माना ${\text{sinx + cosx}}\;{\text{ = }}\;{\text{t}}$

$\begin{array}{*{20}{r}} {{\text{cosx - sinx = dt/dx}}} \\ {{\text{dx = dt/ - (sinx - cosx)}}} \\ {\therefore \;{\text{I = }}\dfrac{1}{2}\left[ {\int {\text{1}} {\text{dx - }}\int {\text{d}} {\text{t/t}}} \right]} \\ {{\text{ = }}\dfrac{1}{2}{\text{[x - log|t|] + C}}} \\ {{\text{ = }}\dfrac{1}{2}{\text{[x - log|sinx + cosx|] + C}}} \end{array}$

33. $\mathbf{{\text{1}}\;{\text{/}}\;({\text{1}}\;{\text{ - }}\;{\text{tanx)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{1}}}{{{\text{(1}}\;{\text{ - }}\;{\text{tanx)}}}}{\text{dx}}}$

\begin{align}{\text{ = }}\int {\dfrac{{\text{1}}}{{{\text{1 - (sinx/cosx)}}}}} {\text{dx}} \hfill \\ {\text{ = }}\int {{\text{(cos}}} {\text{x/(cosx - sinx))dx}} \hfill \\{\text{ = }}\int {\text{2}} {\text{cosx/2(cosx - sinx)dx}} \hfill \\ {\text{ = }}\dfrac{1}{2}\int {{\text{(cosx + cosx + sinx - sinx)}}} {\text{/(cosx - sinx)dx}} \hfill \\ {\text{ = }}\dfrac{1}{2}\left[ {\int {{\text{(cosx - sinx)}}} {\text{/(cosx - sinx)dx + }}} \right.\left. {\int {{\text{(cosx + sinx)}}} {\text{/(cosx - sinx)dx}}} \right] \hfill \\{\text{ = }}\dfrac{1}{2}\left[ {\int {\text{1}} {\text{dx + }}\int {{\text{(cosx + sinx)}}} {\text{/(cosx - sinx)dx}}} \right] \hfill \\ \end{align}

माना ${\text{cosx - sinx}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{ - sinx - cosx = dt/dx}} \hfill \\{\text{ - [sinx + cosx] = dt/dx}} \hfill \\{\text{dx = dt/ - (sinx + cosx)}} \hfill \\\therefore \;{\text{I = }}\dfrac{1}{2}\left[ {\int {\text{1}} {\text{dx + }}\int {{\text{(cos}}} {\text{x + sinx)t}}\dfrac{{{\text{dt}}}}{{{\text{ - (sinx + cosx)}}}}} \right. \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{1}{2}\left[ {\int {\text{1}} {\text{dx - }}\int {\text{d}} {\text{t/t}}} \right] \hfill \\ {\text{ = }}\dfrac{1}{2}{\text{[x - log|t|] + C}} \hfill \\ {\text{ = }}\dfrac{1}{2}{\text{[x - log|cosx - sinx|] + C}} \hfill \\ \end{align}

34. $\mathbf{\sqrt {{\text{tan}}\;{\text{x}}\;{\text{/}}\;{\text{sin}}\;{\text{x}}\;{\text{. cos}}\;{\text{x}}} }$

उत्तर: $\int {{{tanx/(sinx \times cosx)}}} {\text{dx}}$

\begin{align}{\text{ = }}\int {\sqrt {{\text{tanx}}} } {{/(sinxcosx \times cosx/cosx)dx}} \hfill \\{\text{ = }}\int {\sqrt {{{tanx/(tanx \times co}}{{\text{s}}^{\text{2}}}{\text{x)}}} } {\text{ dx}} \hfill \\{\text{ = }}\int {\sqrt {{\text{tanxse}}{{\text{c}}^{\text{2}}}{\text{x/(tanx)}}} \;} {\text{dx}} \hfill \\ \end{align}

माना ${\text{tanx = t}}$

\begin{align}{\text{se}}{{\text{c}}^{\text{2}}}{\text{x = dt/dx}} \hfill \\ {\text{dx = dt/se}}{{\text{c}}^{\text{2}}}{\text{x}} \hfill \\\therefore {\text{ I = }}\int {(\sqrt {\text{t}} } {\text{se}}{{\text{c}}^{\text{2}}}{\text{xdt)/(tse}}{{\text{c}}^{\text{2}}}{\text{x)}} \hfill \\{\text{ = }}\int {{\text{dt}}} {\text{/}}\sqrt {\text{t}} \hfill \\ {\text{ = }}\int {{{\text{t}}^{{\text{ - 1/2}}}}} {\text{dt}} \hfill \\ {\text{ = }}{{\text{t}}^{{\text{( - 1/2) + 1}}}}{\text{/( - 1/2 + 1) + C}} \hfill \\ {\text{ = 2}}\sqrt {\text{t}} {\text{ + C}} \hfill \\{\text{ = 2}}\sqrt {{\text{tanx}}} {\text{ + C}} \hfill \\ \end{align}

35. $\mathbf{{{\text{(1}}\;{\text{ + }}\;{\text{log}}\;{\text{x)}}^{\text{2}}}\;{\text{/}}\;{\text{x}}}$

उत्तर: $\int {{{{\text{(1 + logx)}}}^{\text{2}}}{\text{/x}}} \;{\text{dx}}$

माना       ${\text{1 + logx}}\;{\text{ = }}\;{\text{t}}$

\begin{array}{*{20}{r}}{{\text{1/x = dt/dx}}} \\ {{\text{dx = xdt}}} \\ {{\text{ }}\therefore {\text{ I = }}\int {{\text{(}}{{\text{t}}^{\text{2}}}} {\text{xdt)/(x)}}} \\ \begin{align} {\text{ = }}\int {{{\text{t}}^{\text{2}}}} {\text{dt}} \hfill \\ {\text{ = (}}{{\text{t}}^{\text{3}}}{\text{/3) + C}} \hfill \\ \end{align} \\ {{\text{ = }}\dfrac{{{{{\text{(1 + logx)}}}^{\text{3}}}}}{3}{\text{ + C}}} \end{array}

36. $\mathbf{{\text{(x}}\;{\text{ + }}\;{\text{1)(x}}\;{\text{ + }}\;{\text{log}}\;{\text{x}}{{\text{)}}^{\text{2}}}\;{\text{/}}\;{\text{x}}}$

उत्तर: $\int {\dfrac{{{\text{(x + 1)(x + logx}}{{\text{)}}^{\text{2}}}}}{{\text{x}}}{\text{dx}}}$

\begin{align}{\text{ = }}\;\int {\dfrac{{{\text{(x + 1)}}}}{{\text{x}}}{{{\text{(x + logx)}}}^{\text{2}}}{\text{dx}}} \hfill \\{\text{ = }}\;\int {{\text{(1 + }}\dfrac{{\text{1}}}{{\text{x}}}{\text{)(x + logx}}{{\text{)}}^{\text{2}}}{\text{dx}}} \hfill \\ \end{align}

माना ${\text{x + logx}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{(1 + (1/x))dx}}\;{\text{ = }}\;{\text{dt}} \hfill \\{{\backslash }}\;{\text{I}}\;{\text{ = }}\;\int {{{\text{t}}^{\text{2}}}{\text{dt}}} \hfill \\{\text{ = }}\;{\text{(}}{{\text{t}}^{\text{3}}}{\text{/3) + C}} \hfill \\{\text{ = }}\;{\text{[(x + logx}}{{\text{)}}^{\text{3}}}{\text{/3] + C}} \hfill \\ \end{align}

37. $\mathbf{{{\text{x}}^{\text{3}}}{\text{sin}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\;{{\text{x}}^{\text{4}}}} \right)\;{\text{/}}\;{\text{1 + }}{{\text{x}}^{\text{8}}}}$

उत्तर: $\int {{\text{(}}{{\text{x}}^{\text{3}}}{\text{sin(ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{x}}^{\text{4}}}{\text{))}}} {\text{/(1 + }}{{\text{x}}^{\text{8}}}{\text{)}}\;{\text{dx}}$

माना              ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{x}}^{\text{4}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}\begin{array}{*{20}{c}} {{\text{4}}{{\text{x}}^{\text{3}}}{\text{/(1 + }}{{\text{x}}^{\text{8}}}{\text{) = dt/dx}}} \\ {{\text{dx = }}\left( {{\text{1 + }}{{\text{x}}^{\text{8}}}} \right){\text{dt/4}}{{\text{x}}^{\text{3}}}} \\ {\therefore \;{\text{I = }}\int {{\text{[}}{{\text{x}}^{\text{3}}}} {\text{sint}}\left( {{\text{1 + }}{{\text{x}}^{\text{8}}}} \right){\text{dt]/[}}\left( {{\text{1 + }}{{\text{x}}^{\text{8}}}} \right){\text{4}}{{\text{x}}^{\text{3}}}{\text{]}}} \end{array} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{sin}}} {\text{t dt}} \hfill \\{\text{ = ( - cost/4) + C}} \hfill \\{\text{ = [ - cos}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{{\text{x}}^{\text{4}}}} \right){\text{/4] + C}} \hfill \\ \end{align}

प्रशन 38 एवं 39 में सही उत्तर का चयन कीजिए :

38. $\mathbf{\int {\dfrac{{{\text{(10}}{{\text{x}}^{\text{9}}}{\text{ + 1}}{{\text{0}}^{\text{x}}}{\text{lo}}{{\text{g}}_{\text{e}}}{\text{10}}\;{\text{)}}}}{{{\text{(}}{{\text{x}}^{{\text{10}}}}{\text{ + 1}}{{\text{0}}^{\text{x}}}{\text{)}}}}} \;{\text{dx}}}$ बराबर हैं :

1. $\mathbf{{\text{1}}{{\text{0}}^{\text{x}}}{\text{ - }}{{\text{x}}^{{\text{10}}}}{\text{ + c}}}$

2. $\mathbf{{\text{1}}{{\text{0}}^{\text{x}}}{\text{ + }}{{\text{x}}^{{\text{10}}}}{\text{ + c}}}$

3. $\mathbf{{{\text{(1}}{{\text{0}}^{\text{x}}}{\text{ - }}{{\text{x}}^{{\text{10}}}}{\text{)}}^{{\text{ - 1}}}}{\text{ + c}}}$

4. $\mathbf{{\text{log(1}}{{\text{0}}^{\text{x}}}{\text{ + }}{{\text{x}}^{{\text{10}}}}{\text{) + c}}}$

उत्तर: $\int {\dfrac{{{\text{(10}}{{\text{x}}^{\text{9}}}{\text{ + 1}}{{\text{0}}^{\text{x}}}{\text{lo}}{{\text{g}}_{\text{e}}}{\text{10}}\;{\text{)}}}}{{{\text{(}}{{\text{x}}^{{\text{10}}}}{\text{ + 1}}{{\text{0}}^{\text{x}}}{\text{)}}}}} \;{\text{dx}}$

माना ${{\text{x}}^{{\text{10}}}}{\text{ + 10x = t}}$

\begin{align}\left( {{\text{10}}{{\text{x}}^{\text{9}}}{{ + 10 \times lo}}{{\text{g}}_{\text{e}}}{\text{10}}} \right){\text{dx = dt}} \hfill \\{\text{dx = dt/}}\left( {{\text{10}}{{\text{x}}^{\text{9}}}{\text{ + 10x lo}}{{\text{g}}_{\text{e}}}} \right) \hfill \\\therefore \;{\text{I = }}\int {{\text{[}}\left( {{\text{10}}{{\text{x}}^{\text{9}}}{{ + 10 \times lo}}{{\text{g}}_{\text{e}}}{\text{10}}} \right)} {\text{dt]/[t}}\left( {{\text{10}}{{\text{x}}^{\text{9}}}{{ + 10 \times lo}}{{\text{g}}_{\text{e}}}{\text{10}}} \right){\text{]}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\int {{\text{dt/t}}} \hfill \\{\text{ = log}}\left| {\text{t}} \right|{\text{ + C}} \hfill \\ \end{align}

${\text{ = log}}\left| {{\text{1}}{{\text{0}}^{\text{x}}}{\text{ + }}{{\text{x}}^{{\text{10}}}}} \right|{\text{ + C}}$

विकल्प (d) सही है |

39.$\mathbf{\dfrac{{{\text{dx}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}\;{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}}$ बराबर हैं:

1. $\mathbf{{\text{tan}}\;{\text{x}}\;{\text{ + }}\;{\text{cot}}\;{\text{x + }}\;{\text{c }}}$

2. $\mathbf{{\text{tan}}\;{\text{x}}\;{\text{ - }}\;{\text{cot}}\;{\text{x + }}\;{\text{c }}}$

3. $\mathbf{{\text{tan}}\;{\text{x}}\;{\text{cot}}\;{\text{x + }}\;{\text{c }}}$

4. $\mathbf{{\text{tan}}\;{\text{x}}\;{\text{ - }}\;{\text{cot}}\;{\text{2x + }}\;{\text{c }}}$

उत्तर: $\int {\dfrac{{{\text{dx}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}\;{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}}$

\begin{align}{\text{ = }}\int {\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} {\text{/(si}}{{\text{n}}^{\text{2}}}{{x \times co}}{{\text{s}}^{\text{2}}}{\text{x)dx}} \hfill \\{\text{ = }}\int {{\text{si}}{{\text{n}}^{\text{2}}}}{\text{x/(si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x)dx + }}\int {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{x/(si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x)dx}} \hfill \\{\text{ = }}\int {{\text{(1}}} {\text{/co}}{{\text{s}}^{\text{2}}}{\text{x)dx + }}\int {{\text{(1}}} {\text{/si}}{{\text{n}}^{\text{2}}}{\text{x)dx}} \hfill \\{\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{x dx + }}\int {{\text{cose}}{{\text{c}}^{\text{2}}}} {\text{x dx}} \hfill \\ {\text{ = tanx - cotx + C}} \hfill \\ \end{align}

विकल्प (b) सही है |

### प्रश्नावली 7.3

1 से 22 तक के प्रश्नों मे प्रत्येक फलन का समाकलन ज्ञात कीजिए:

1. $\mathbf{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(2x + 5)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^2}(2x + 5)$

\begin{align}\int {\dfrac{{{\text{1 - cos(2x + 5)}}}}{{\text{2}}}} {\text{dx}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{[1 - cos(4x + 10)]dx}}} \hfill \\{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{dx}}} \;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{cos(4x + 10)dx}}} \hfill \\ \end{align}

माना की ${\text{4x + 10}}\;{\text{ = }}\;{\text{t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{4dx}}\;{\text{ = }}\;{\text{dt}} \hfill \\{\text{dx}}\;{\text{ = }}\;\dfrac{{{\text{dt}}}}{{\text{4}}} \hfill \\{\text{I}}\;{\text{ = }}\;\dfrac{{\text{x}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{cos(t)}}} {\text{dt}} \hfill \\{\text{ = }}\;\dfrac{{\text{x}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{sint + C}} \hfill \\{\text{ = }}\;\dfrac{{\text{x}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{sin(4x + 10) + C}} \hfill \\ \end{align}

2. $\mathbf{{\text{sin3x cos4x}}}$

उत्तर: $\int {{\text{sin3x}}\;{\text{cos4x}}\;{\text{dx}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\text{2}} {\text{ sin3x cos4x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{(sin7x - sinx)}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{\text{ - }}\dfrac{{{\text{cos7x}}}}{{\text{7}}}{\text{ + cosx}}} \right){\text{ + C}} \hfill \\ \end{align}

3. $\mathbf{{\text{cos2x cos4x cos6x}}}$

उत्तर: $\int {{\text{cos2x}}\;{\text{cos4x}}\;{\text{cos6x}}\;{\text{dx}}}$

\begin{align}= \;\dfrac{1}{2}\int {\text{2}} {\text{ cos2x cos4x cos6x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{(cos6x + cos2x)}}} {\text{cos6x dx}} \hfill \\ \end{align}\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{(6x) + cos2x cos6x}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\left( {{\text{2co}}{{\text{s}}^{\text{2}}}{\text{(6x)}}} \right)} {\text{dx + }}\dfrac{{\text{1}}}{{\text{4}}}\int {\text{2}} {\text{ cos2x cos6x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{(1 + cos12x)}}} {\text{dx + }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{(cos}}} {\text{8x + cos4x)dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}{\text{x + }}\dfrac{{{\text{sin12x}}}}{{{\text{48}}}}{\text{ + }}\dfrac{{{\text{sin8x}}}}{{{\text{32}}}}{\text{ + }}\dfrac{{{\text{sin4x}}}}{{{\text{16}}}}{\text{ + C}} \hfill \\ \end{align}

4. $\mathbf{{\text{si}}{{\text{n}}^{\text{3}}}{\text{(2x + 1)}}}$

उत्तर: $\int {{\text{si}}{{\text{n}}^{\text{3}}}{\text{(2x + 1)dx}}}$

\begin{align}{\text{ = }}\;\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{[3sin(2x + 1) - sin(2x + 1)}}} {\text{]dx}} \hfill \\{\text{ = }}\;\dfrac{{\text{3}}}{{\text{4}}}{\text{( - }}\dfrac{{{\text{cos(2x + 1)}}}}{{\text{2}}}{\text{) - }}\dfrac{{\text{3}}}{{\text{4}}}{\text{( - }}\dfrac{{{\text{cos3(2x + 1)}}}}{{\text{6}}}{\text{) + C}} \hfill \\{\text{ = }}\;{\text{ - }}\dfrac{{\text{3}}}{{\text{8}}}{\text{cos(2x + 1) + }}\dfrac{{\text{1}}}{{{\text{24}}}}{\text{cos3(2x + 1) + C}} \hfill \\{\text{ = }}\;{\text{ - }}\dfrac{{\text{3}}}{{\text{8}}}{\text{cos(2x + 1) + }}\dfrac{{\text{1}}}{{{\text{24}}}}{\text{[4co}}{{\text{s}}^{\text{3}}}{\text{(2x + 1) - 3cos(2x + 1)] + C}} \hfill \\{\text{ = }}\;{\text{ - }}\dfrac{{\text{3}}}{{\text{8}}}{\text{cos(2x + 1) + }}\dfrac{{\text{1}}}{{\text{6}}}{\text{[4co}}{{\text{s}}^{\text{3}}}{\text{(2x + 1) - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{cos(2x + 1)] + C}} \hfill \\{\text{ = }}\;{\text{ - }}\dfrac{{\text{3}}}{{\text{8}}}{\text{cos(2x + 1) + }}\dfrac{{\text{1}}}{{\text{6}}}{\text{[co}}{{\text{s}}^{\text{3}}}{\text{(2x + 1)] + C}} \hfill \\ \end{align}

5. $\mathbf{{\text{si}}{{\text{n}}^{\text{3}}}{\text{(x) co}}{{\text{s}}^{\text{3}}}{\text{(x)}}}$

उत्तर: $\int {{\text{si}}{{\text{n}}^{\text{3}}}{\text{x}}\;{\text{co}}{{\text{s}}^{\text{3}}}{\text{x}}\;{\text{dx}}}$

\begin{align}{\text{ = }}\int {{\text{sin}}} {\text{x si}}{{\text{n}}^{\text{2}}}{\text{(x) co}}{{\text{s}}^{\text{3}}}{\text{(x) dx}} \hfill \\{\text{ = }}\int {{\text{sin}}} {\text{x}}\left( {{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{(x)}}} \right){\text{co}}{{\text{s}}^{\text{3}}}{\text{(x) dx}} \hfill \\{\text{ = }}\int {\left( {{\text{co}}{{\text{s}}^{\text{3}}}{\text{(x) - co}}{{\text{s}}^{\text{5}}}{\text{(x)}}} \right)} {\text{sinx dx}} \hfill \\ \end{align}

माना की ${\text{cosx}}\;{\text{ = }}\;{\text{t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{ - sinx}}\;{\text{dx}}\;{\text{ = }}\;{\text{dt }} \Rightarrow \;{\text{sinx}}\;{\text{dx}}\;{\text{ = }}\;{\text{ - dt}} \hfill \\ \int {{\text{(}}{{\text{t}}^{\text{3}}}{\text{ - }}{{\text{t}}^{\text{5}}}{\text{)( - dt)}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{{\text{t}}^{\text{4}}}}}{{\text{4}}}{\text{ + }}\dfrac{{{{\text{t}}^{\text{6}}}}}{{\text{6}}}{\text{ + C}}} \hfill \\{\text{ = }}\;{\text{(1/6)co}}{{\text{s}}^{\text{6}}}{\text{x - (1/4)co}}{{\text{s}}^{\text{4}}}{\text{x + C}} \hfill \\ \end{align}

6. $\mathbf{{\text{sinx sin2x sin3x }}}$

उत्तर: $\int {{\text{sinx}}\;{\text{sin2x}}\;{\text{sin3x}}\;{\text{dx}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{2 }}} {\text{sinx sin2x sin3x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{(sinx - sin3x)}}} {\text{sin3x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{2 }}} {\text{sin3x cosx dx - }}\dfrac{{\text{1}}}{{\text{4}}}\int {\text{2}} {\text{ sin3x cos3x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{(sin4x + sin2x)}}} {\text{dx - }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{sin}}} {\text{6x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{16}}}}{\text{cos4x - }}\dfrac{{\text{1}}}{{\text{8}}}{\text{cos2x + }}\dfrac{{\text{1}}}{{{\text{24}}}}{\text{cos6x + C}} \hfill \\ \end{align}

7. $\mathbf{{\text{sin4x sin8x}}}$

उत्तर: $\int {{\text{sin4x}}\;{\text{sin8x}}\;{\text{dx}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{2 }}} {\text{sin4x sin8x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{(cos(}}} {\text{4x) - sin(12x))sin3x dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{{\text{sin4x}}}}{{\text{4}}}{\text{ - }}\dfrac{{{\text{sin12x}}}}{{{\text{12}}}}} \right]{\text{ + C}} \hfill \\ \end{align}

8. $\mathbf{\dfrac{{{\text{1 - cosx}}}}{{{\text{1 + cosx}}}}}$

उत्तर: $\int {\dfrac{{{\text{1 - cosx}}}}{{{\text{1 + cosx}}}}{\text{dx}}}$

\begin{align}{\text{ = }}\int{\dfrac{{{\text{2si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ 2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} \left( {\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{dx}} \hfill \\{\text{ = }}\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ - 1}}} \right)} {\text{ dx}} \hfill \\{\text{ = 2tan}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - x + C}} \hfill \\ \end{align}

9. $\mathbf{\dfrac{{{\text{cosx}}}}{{{\text{1 + cosx}}}}}$

उत्तर: $\int {\dfrac{{{\text{cosx}}}}{{{\text{1 + cosx}}}}} {\text{ dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{{\text{1 + cosx - 1}}}}{{{\text{1 + cosx}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {\text{1}} {\text{ dx + }}\int {\dfrac{{\text{1}}}{{{\text{1 + cosx}}}}} {\text{ dx}} \hfill \\{\text{ = x - }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{se}}{{\text{c}}^{\text{2}}}} \left( {\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{dx + C}} \hfill \\ \end{align}

${\text{ = x - tan}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + C}}$

10. $\mathbf{{\text{si}}{{\text{n}}^{\text{4}}}{\text{(x)}}}$

उत्तर: $\int {{\text{si}}{{\text{n}}^{\text{4}}}} {\text{(x)dx}}$

\begin{align}{\text{ = }}\int {{{\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)}}} \right)}^{\text{2}}}} {\text{dx}} \hfill \\{\text{ = }}\int {{{\left( {\dfrac{{{\text{1 - cos2x}}}}{{\text{2}}}} \right)}^{\text{2}}}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\left( {{\text{1 + co}}{{\text{s}}^{\text{2}}}{\text{(2x) - 2cos2x}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\left( {{\text{1 + }}\dfrac{{{\text{1 + cos4x}}}}{{\text{2}}}{\text{ - 2cos2x}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {{\text{(1) }}} {\text{dx + }}\dfrac{{\text{1}}}{{\text{8}}}\int {{\text{(1 + cos4x)}}} {\text{ dx - }}\dfrac{{\text{2}}}{{\text{4}}}\int {{\text{(cos2x)}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{x}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{x}}}{{\text{8}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{8}}}{{ \times }}\dfrac{{{\text{sin4x}}}}{{\text{4}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times }}\dfrac{{{\text{sin2x}}}}{{\text{2}}} \hfill \\{\text{ = }}\dfrac{{\text{3}}}{{\text{8}}}{\text{x + }}\dfrac{{\text{1}}}{{{\text{32}}}}{\text{sin4x - }}\dfrac{{\text{1}}}{{\text{4}}}{\text{sin2x + C}} \hfill \\ \end{align}

11. ${\text{co}}{{\text{s}}^{\text{4}}}{\text{(2x)}}$

उत्तर: $\int {{\text{co}}{{\text{s}}^{\text{4}}}} {\text{(2x)dx}}$

\begin{align}{\text{ = }}\int {{{\left( {\dfrac{{{\text{1 + cos4x}}}}{{\text{2}}}} \right)}^{\text{2}}}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\left( {{\text{1 + co}}{{\text{s}}^{\text{2}}}{\text{(4x) + 2cos4x}}} \right)} {\text{ dx}} \hfill \\ \end{align}

${\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\left( {{\text{1 + }}\dfrac{{{\text{1 + cos8x}}}}{{\text{2}}}{\text{ + 2cos4x}}} \right)} {\text{ dx}}$

\begin{align}{\text{ = }}\dfrac{{\text{x}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{x}}}{{\text{8}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{8}}}{{ \times }}\dfrac{{{\text{sin8x}}}}{{\text{8}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times }}\dfrac{{{\text{sin4x}}}}{{\text{4}}} \hfill \\{\text{ = }}\dfrac{{\text{3}}}{{\text{8}}}{\text{x + }}\dfrac{{\text{1}}}{{{\text{64}}}}{\text{sin8x + }}\dfrac{{\text{1}}}{{\text{8}}}{\text{sin4x + C}} \hfill \\ \end{align}

12. $\mathbf{\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{1 + cosx}}}}}$

उत्तर: $\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{1 + cosx}}}}} {\text{ dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{\text{1 + cosx}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(1 - cosx)(1 + cosx)}}}}{{{\text{1 + cosx}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {{\text{(1 - cosx) }}} {\text{dx}} \hfill \\{\text{ = x - sinx + C}} \hfill \\ \end{align}

13. $\mathbf{\dfrac{{{{cos2x - cos2\alpha }}}}{{{{cosx - cos\alpha }}}}}$

उत्तर: $\int {\dfrac{{{{cos2x - cos2\alpha }}}}{{{{cosx - cos\alpha }}}}} {\text{ dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{\left( {{\text{2co}}{{\text{s}}^{\text{2}}}{\text{(x) - 1}}} \right){\text{ - }}\left( {{\text{2co}}{{\text{s}}^{\text{2}}}{{(\alpha ) - 1}}} \right)}}{{{{cosx - cos\alpha }}}}} {\text{ dx}} \hfill \\ \because \;{\text{cos2x = 2co}}{{\text{s}}^{\text{2}}}{\text{(x) - 1}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{2}}\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x) - co}}{{\text{s}}^{\text{2}}}{{(\alpha )}}} \right){\text{ - 1 + 1}}}}{{{{cosx - cos\alpha }}}}} \hfill \\{\text{ = }}\int {\dfrac{{{{2(cosx - cos\alpha )(cosx + cos\alpha )}}}}{{{{cosx - cos\alpha }}}}} \hfill \\{\text{ = }}\int {\text{2}} {{(cosx - cos\alpha )}}\;{\text{dx}} \hfill \\ \end{align}

${\text{ = 2(sinx - xcos\alpha ) + C}}$

14. $\mathbf{\dfrac{{{\text{cosx - sinx}}}}{{{\text{1 + sin2x}}}}}$

उत्तर: $\int {\dfrac{{{\text{cosx - sinx}}}}{{{\text{1 + sin2x}}}}} {\text{dx}}$

${\text{ = }}\int {\dfrac{{{\text{(cosx - sinx)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x) + si}}{{\text{n}}^{\text{2}}}{\text{(x) + 2sinxcosx}}}}} {\text{ dx}}$

\begin{align}\because \;{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x) + si}}{{\text{n}}^{\text{2}}}{\text{(x) = 1}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(cosx - sinx)}}}}{{{{{\text{(cosx + sinx)}}}^{\text{2}}}}}} {\text{ dx}} \hfill \\ \end{align}

माना की ${\text{cosx + sinx = t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{( - sin(x) + cosx)dx = dt, sin(x)dx = - dt}} \hfill \\\int {\dfrac{{\text{1}}}{{{{\text{t}}^{\text{2}}}}}\;{\text{dt}}} \hfill \\{\text{ = - }}\dfrac{{\text{1}}}{{\text{t}}}{\text{ + C}} \hfill \\{\text{ = -}}\dfrac{{\text{1}}}{{{\text{(cosx + sinx)}}}}{\text{ + C}} \hfill \\ \end{align}

15. $\mathbf{{\text{ta}}{{\text{n}}^{\text{3}}}{\text{(2x) sec2x}}}$

उत्तर: $\int {{\text{ta}}{{\text{n}}^{\text{3}}}{\text{(2x}}} {\text{)}}\;{\text{sec(2x)}}\;{\text{dx}}$

$\int {{\text{tan(2x)}}\;{\text{ta}}{{\text{n}}^2}{\text{(2x}}} {\text{)}}\;{\text{sec(2x)}}\;{\text{dx}}$

$\int {{\text{(se}}{{\text{c}}^{\text{2}}}{\text{(2x)}}} {\text{ - 1)}}\;{\text{sec(2x)}}\;{\text{tan(2x)}}\;{\text{dx}}$

माना की ${\text{sec2x = t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{2 sec2x tan2x dx = dt}} \hfill \\{\text{sec2x tan2x dx = }}\dfrac{{{\text{dt}}}}{{\text{2}}} \hfill \\{\text{ = }}\int {\left( {{{\text{t}}^{\text{2}}}{\text{ - 1}}} \right)} \dfrac{{{\text{dt}}}}{{\text{2}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{{{\text{t}}^{\text{3}}}}}{{\text{3}}}{\text{ - t}}} \right) \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{{\text{se}}{{\text{c}}^{\text{3}}}{\text{2x}}}}{{\text{3}}}{\text{ - sec2x}}} \right){\text{ + C}} \hfill \\ \end{align}

16. $\mathbf{{\text{ta}}{{\text{n}}^{\text{4}}}{\text{(x)}}}$

उत्तर: $\int {{\text{ta}}{{\text{n}}^{\text{4}}}{\text{(x)}}\;{\text{dx}}}$

\begin{align}{\text{ = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(x)ta}}{{\text{n}}^{\text{2}}}{\text{(x) dx}} \hfill \\{\text{ = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(x)}}\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{(x) - 1}}} \right){\text{dx}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(x)se}}{{\text{c}}^{\text{2}}}{\text{(x)dx - }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(x)dx}} \hfill \\{\text{ = }}\int {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{(x)se}}{{\text{c}}^{\text{2}}}{\text{(x)dx - }}\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{(x) - 1}}} \right)} {\text{dx}} \hfill \\ \end{align}

माना की ${\text{tanx = t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

${\text{se}}{{\text{c}}^{\text{2}}}{\text{(x)dx = dt}}$

\begin{align}{\text{ = }}\int {\left( {{{\text{t}}^{\text{2}}}} \right)} {\text{dt - }}\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{(x) - 1}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{t}}^{\text{3}}}}}{{\text{3}}}{\text{ - (tanx - x) + C}} \hfill \\{\text{ = }}\;\dfrac{{{\text{ta}}{{\text{n}}^{\text{3}}}{\text{x}}}}{{\text{3}}}{\text{tanx + x + C}} \hfill \\ \end{align}

17.$\mathbf{\dfrac{{\text{si}}{{\text{n}}^{\text{3}}}{\text{(x)-co}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}$

उत्तर: $\int {\left( {\dfrac{{{\text{si}}{{\text{n}}^{\text{3}}}{\text{(x) - co}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} \right)} {\text{ dx}}$

\begin{align}{\text{ = }}\int {\left( {\dfrac{{{\text{si}}{{\text{n}}^{\text{3}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} \right)} {\text{ - }}\left( {\dfrac{{{\text{co}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} \right){\text{ dx}} \hfill \\{\text{ = }}\int {\left( {\dfrac{{{\text{sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} \right)} {\text{ - }}\left( {\dfrac{{{\text{cosx}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}} \right){\text{dx}} \hfill \\{\text{ = }}\int {{\text{[(sec}}} {\text{x tanx) + (cosecx cotx)] dx}} \hfill \\{\text{ = }}\int {{\text{(sec}}} {\text{x tanx) dx + }}\int {{\text{(cosec}}} {\text{x cotx) dx}} \hfill \\ {\text{ = secx - cosecx}} \hfill \\ \end{align}

18. $\mathbf{\dfrac{{{\text{cos2x + 2si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}}$

उत्तर: $\int {\dfrac{{{\text{cos2x + 2si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} {\text{dx}}$

\begin{array}{*{20}{l}}{{\text{ = }}\int {\dfrac{{{\text{1 - 2si}}{{\text{n}}^{\text{2}}}{\text{(x) + 2si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}}{\text{dx}}} \\ {{\text{ = }}\int {\dfrac{{\text{1}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}} {\text{dx}}} \\ \begin{align}{\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{(x)dx}} \hfill \\{\text{ = tanx + C}} \hfill \\ \end{align} \end{array}

19.$\mathbf{\dfrac{{{\text{cos2x}}}}{{{\text{sinxco}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}}$

उत्तर: $\int {\dfrac{{\text{1}}}{{{\text{sinxco}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}} {\text{ dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x) + co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{\text{sinxco}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {\left( {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)}}}}{{{\text{sinxco}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}{\text{+}}\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{\text{sinxco}}{{\text{s}}^{\text{3}}}{\text{(x)}}}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\int {\left( {{\text{tanx se}}{{\text{c}}^{\text{2}}}{\text{(x) + }}\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{(x)}}}}{{{\text{tanx}}}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\int {\left( {{\text{tanx + }}\dfrac{{\text{1}}}{{{\text{tanx}}}}} \right)} {\text{ se}}{{\text{c}}^{\text{2}}}{\text{(x) dx}} \hfill \\ \end{align}

माना की ${\text{tanx = t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{se}}{{\text{c}}^{\text{2}}}{\text{(x)dx = dt}} \hfill \\{\text{ = }}\int {\left( {{\text{t + }}\dfrac{{\text{1}}}{{\text{t}}}} \right)} {\text{ dt}} \hfill \\ {\text{ = }}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{3}}}{\text{ + logt + C}} \hfill \\{\text{ = }}\dfrac{{{\text{ta}}{{\text{n}}^{\text{2}}}{\text{x}}}}{{\text{2}}}{\text{ + log[tanx] + C}} \hfill \\ \end{align}

20. $\mathbf{\dfrac{{{\text{cos2x}}}}{{{{{\text{(cosx + sinx)}}}^{\text{2}}}}}}$

उत्तर: $\int {\dfrac{{{\text{cos2x}}}}{{{{{\text{(cosx + sinx)}}}^{\text{2}}}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x) - co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{{{\text{(cosx + sinx)}}}^{\text{2}}}}}} {\text{dx}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(cosx - sinx)(cosx + sinx)}}}}{{{{{\text{(cosx + sinx)}}}^{\text{2}}}}}} {\text{dx}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(cosx - sinx)}}}}{{{\text{(cosx + sinx)}}}}} {\text{dx}} \hfill \\ \end{align}

माना की ${\text{cosx + sinx = t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{\text{( - sin(x) + cosx)dx = dt}} \hfill \\{\text{sin(x)dx = - dt}} \hfill \\{\text{ = }}\int {\left( {\dfrac{{\text{1}}}{{\text{t}}}} \right)} {\text{dt}} \hfill \\{\text{ = logt + C}} \hfill \\{\text{ = log[cosx + sinx] + C}} \hfill \\ \end{align}

21. $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(cosx)}}}$

उत्तर: $\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{(cosx)dx}}$

\begin{align}{\text{ = }}\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{(sin(}}\dfrac{\pi }{2}{\text{ - x))dx}} \hfill \\{\text{ = }}\int {\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{x - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + C}} \hfill \\ \end{align}

22. $\mathbf{\dfrac{{\text{1}}}{{{\text{cos(x - a)cos(x - b)}}}}}$

उत्तर: $\int {\dfrac{{\text{1}}}{{{\text{cos(x - a)cos(x - b)}}}}} {\text{dx}}$

${\text{ = sin(a - b)}}$ से अंश तथा हर मे गुना करने पर

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}\int {\dfrac{{{\text{sin(a - b)}}}}{{{\text{cos(x - a)cos(x - b)}}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}\int {\dfrac{{{\text{sin[(x - b) - (x - a)]}}}}{{{\text{cos(x - a)cos(x - b)}}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}\int {\dfrac{{{\text{sin(x - b)cos(x - a) - cos(x - b)sin(x - a)}}}}{{{\text{cos(x - a)cos(x - b)}}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}\left[ {\int {{\text{tan}}} {\text{(x - b)dx - }}\int {{\text{tan}}} {\text{(x - a)dx}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}{\text{[ - log|cos(x - b)| + log|cos(x - a)|] + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{sin(a - b)}}}}\left[ {{\text{ - log}}\left| {\dfrac{{{\text{cos(x - a)}}}}{{{\text{cos(x - b)}}}}} \right|} \right]{\text{ + C}} \hfill \\ \end{align}

### 23 और 24 मे सही उत्तर का चयन कीजिए:

23. $\mathbf{\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x) - co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{\text{dx}}} }$ बराबर है:

1. $\mathbf{{\text{tanx + cotx + C}}}$

2. $\mathbf{{\text{tanx + cosecx + C}}}$

3. $\mathbf{{\text{ - tanx + cotx + C}}}$

4. $\mathbf{{\text{tanx + secx + C}}}$

उत्तर: $\int {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x) - co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}{\text{(x)co}}{{\text{s}}^{\text{2}}}{\text{(x)}}}}{\text{dx}}}$

\begin{align}{\text{ = }}\int {{\text{[se}}{{\text{c}}^{\text{2}}}} {\text{(x) - cose}}{{\text{c}}^{\text{2}}}{\text{(x)]dx}} \hfill \\{\text{ = tanx + cotx + C}} \hfill \\ \end{align}

अतः विकल्प (a) सही है।

24. $\mathbf{\int {\dfrac{{{{\text{e}}^{\text{x}}}{\text{(1 + x)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}\left( {{{\text{e}}^{\text{x}}}{\text{x}}} \right)}}} {\text{dx}}}$ बराबर है:

1. $\mathbf{{\text{ - cot }}{{\text{e}}^{\text{x}}}{\text{x + C}}}$

2. $\mathbf{{\text{tanx }}{{\text{e}}^{\text{x}}}{\text{ + C}}}$

3. $\mathbf{{\text{tan(}}{{\text{e}}^{\text{x}}}{\text{) + C}}}$

4. $\mathbf{{\text{cot(}}{{\text{e}}^{\text{x}}}{\text{) + C}}}$

उत्तर: $\int {\dfrac{{{{\text{e}}^{\text{x}}}{\text{(1 + x)}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}\left( {{{\text{e}}^{\text{x}}}{\text{x}}} \right)}}} {\text{dx}}$

माना की ${{\text{e}}^{\text{x}}}{\text{x}}\;{\text{ = }}\;{\text{t}}$

दोनों पक्षों का ${\text{t}}$ के सापेक्ष अवकलन करने पर

\begin{align}{{\text{e}}^{\text{x}}}{\text{(1 + x)dx = dt}} \hfill \\{\text{ = }}\int {\dfrac{{\text{1}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{(t)}}}}} {\text{dt}} \hfill \\{\text{ = }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{(t)dt}} \hfill \\{\text{ = tant + C}} \hfill \\{\text{ = tan}}\left( {{{\text{e}}^{\text{x}}}{\text{x}}} \right){\text{ + C}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

### 1 से 23 तक के फलनों का समाकलन कीजिए:

1. $\mathbf{\dfrac{{{\text{3}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{6}}}{\text{ + 1}}}}}$

उत्तर: $\int {\dfrac{{{\text{3}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{6}}}{\text{ + 1}}}}{\text{dx}}}$

माना ${\text{u = }}{{\text{x}}^3}$

\begin{align}{\text{du = 3}}{{\text{x}}^{\text{2}}}{{ \times dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{du}}}}{{{{\text{u}}^{\text{2}}}{\text{ + 1}}}}} {\text{ = }}\int {\dfrac{{{\text{du}}}}{{{{\text{u}}^{\text{2}}}{\text{ + (1}}{{\text{)}}^{\text{2}}}}}} \hfill \\\left\{ {\because \;\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = ta}}{{\text{n}}^{ - 1}}{\text{(u) + c}}\quad {\text{\{ y = u, a = 1\} }} \hfill \\ \end{align}

${\text{I = ta}}{{\text{n}}^{ - 1}}{\text{(}}{{\text{x}}^3}{\text{) + c}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{1 + 4}}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: $\int {\dfrac{{\text{1}}}{{\sqrt {{\text{1 + 4}}{{\text{x}}^{\text{2}}}} }}{\text{dx}}}$

माना ${\text{u}}\;{\text{ = }}\;{\text{2x}}$

\begin{align}{\text{du = 2dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{\text{1 + }}{{\text{u}}^{\text{2}}}} }}} \hfill \\\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log|y + }}\sqrt {{{\text{y}}^2}{\text{ + }}{{\text{a}}^2}} {\text{| + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + 1}}} } \right|{\text{ + c \{ y = u, a = 1\} }} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{\text{2x + }}\sqrt {{\text{4}}{{\text{x}}^{\text{2}}}{\text{ + 1}}} } \right|{\text{ + c}} \hfill \\ \end{align}

3. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{{{\text{(2 - x)}}}^{\text{2}}}{\text{ + 1}}} }}}$

उत्तर: $\int {\dfrac{{\text{1}}}{{\sqrt {{{{\text{(2 - x)}}}^{\text{2}}}{\text{ + 1}}} }}{\text{dx}}}$

माना ${\text{u}}\;{\text{ = }}\;{\text{2 - x}}$

\begin{align}{\text{du}}\;{\text{ = }}\;{\text{ - dx}} \hfill \\{\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{ - du}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + 1}}} }}} \hfill \\ \end{align}

\begin{align}\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{\text{I = - log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + 1}}} } \right|{\text{ + c}}\quad {\text{\{ y = u, a = 1\} }} \hfill \\{\text{I = - log}}\left| {{\text{(2 - x) + }}\sqrt {{{{\text{(2 - x)}}}^{\text{2}}}{\text{ + 1}}} } \right|{\text{ + c}} \hfill \\{\text{I = log}}\left| {\dfrac{{\text{1}}}{{{\text{2 - x + }}{{\sqrt {\text{x}} }^{\text{2}}}{\text{ - 4x + 5}}}}} \right|{\text{ + c}} \hfill \\ \end{align}

4. $\mathbf{\dfrac{{\text{1}}}{{\sqrt {{\text{9 - 25}}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: $\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{9 - 25}}{{\text{x}}^{\text{2}}}} }}} \; = \;{\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{3}}^2}{\text{ - (5x}}{{\text{)}}^{\text{2}}}} }}}$

माना ${\text{u}}\;{\text{ = }}\;{\text{5x}}$

\begin{align}{\text{du = 5dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{5}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{{\text{(3)}}}^{\text{2}}}{\text{ - (u}}{{\text{)}}^{\text{2}}}} }}} \hfill \\\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}} {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{5}}}{\text{si}}{{\text{n}}^{{\text{-1}}}}\left({\dfrac{{\text{u}}}{{\text{3}}}} \right){\text{ + c}}\quad {\text{\{ y = u, a = 3\} }} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{5}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{5x}}}}{{\text{3}}}} \right){\text{ + c}} \hfill \\ \end{align}

5. $\mathbf{\dfrac{{{\text{3x}}}}{{{\text{1 + 2}}{{\text{x}}^{\text{4}}}}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{3xdx}}}}{{{\text{1 + 2}}{{\text{x}}^{\text{4}}}}}}$

माना ${\text{u = }}{{\text{x}}^2}$

\begin{align}{\text{du = 2x dx}} \hfill \\{\text{I = }}\dfrac{{\text{3}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{{\text{1 + 2}}{{\text{u}}^{\text{2}}}}}} \hfill \\ \end{align}

${\text{ = }}\dfrac{{\text{3}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{{\text{1 + (}}\sqrt {\text{2}} {\text{u}}{{\text{)}}^{\text{2}}}}}}$

माना $\sqrt {\text{2}} {\text{u = w}}$

\begin{align}\sqrt {\text{2}} {\text{du = dw}} \hfill \\{\text{I = }}\dfrac{{\text{3}}}{{{\text{2}}\sqrt {\text{2}} }}\int {\dfrac{{{\text{dw}}}}{{{\text{1 + }}{{\text{w}}^{\text{2}}}}}} \hfill \\\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{3}}}{{{\text{2}}\sqrt {\text{2}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(w) + c \{ y = w, a = 1\} }} \hfill \\{\text{I = }}\dfrac{{\text{3}}}{{{\text{2}}\sqrt {\text{2}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(}}\sqrt {\text{2}} {\text{u) + c}} \hfill \\{\text{I = }}\dfrac{{\text{3}}}{{{\text{2}}\sqrt {\text{2}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\sqrt {\text{2}} {{\text{x}}^{\text{2}}}} \right){\text{ + c}} \hfill \\ \end{align}

6. $\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{6}}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{1 - }}{{\text{x}}^{\text{6}}}}}{\text{dx}}}$

माना ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{3}}}$

\begin{align}{\text{du}}\;{\text{ = }}\;{\text{3}}{{\text{x}}^{\text{2}}}{\text{dx}} \hfill \\{\text{I}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{3}}}\int {\dfrac{{{\text{du}}}}{{{\text{1 - }}{{\text{u}}^{\text{2}}}}}} \hfill \\ \end{align}

\begin{align}\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{{\text{2a}}}}{\text{log}}\left| {\dfrac{{{\text{a + y}}}}{{{\text{a - y}}}}} \right|{\text{ + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{log}}\left| {\dfrac{{{\text{1 + u}}}}{{{\text{1 - u}}}}} \right|{\text{ + c}}\quad {\text{\{ y = u, a = 1\} }} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{log}}\left| {\dfrac{{{\text{1 + }}{{\text{x}}^{\text{3}}}}}{{{\text{1 - }}{{\text{x}}^{\text{3}}}}}} \right|{\text{ + c}} \hfill \\\end{align}

7. $\dfrac{{{\text{x - 1}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{x - 1}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx}}$

${\text{I}}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx - }}\int {\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx}}$

माना ${{\text{I}}_1}\;{\text{ = }}\;\int {\dfrac{{\text{x}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx , }}{{\text{I}}_2}\; = \;\int {\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx}}$

\begin{align} \therefore \;{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ - }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{xdx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} \hfill \\ \end{align}

माना ${\text{u = }}{{\text{x}}^2}{\text{ - 1}}$

${\text{du = 2x dx}}$

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ \times 2}}\sqrt {\text{u}} {\text{ + c}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\sqrt {\text{u}} {\text{ + c}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\sqrt {{{\text{x}}^2}{\text{ - 1}}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} \;\;\,\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{\text{y}}{\quad ^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = log|x + }}\sqrt {{{\text{x}}^2}{\text{ - 1}}} {\text{| + c}}\quad {\text{\{ y = x, a = 1\} }} \hfill \\ {\text{I = }}{{\text{I}}_1} - {{\text{I}}_2} \hfill \\{\text{I = }}\sqrt {{{\text{x}}^2}{\text{ - 1}}} {\text{ - log|x + }}\sqrt {{{\text{x}}^2}{\text{ - 1}}} {\text{| + c}} \hfill \\ \end{align}

8. $\dfrac{{{{\text{x}}^{\text{2}}}}}{{\sqrt {{{\text{x}}^{\text{6}}}{\text{ + }}{{\text{a}}^{\text{6}}}} }}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\sqrt {{{\text{x}}^{\text{6}}}{\text{ + }}{{\text{a}}^{\text{6}}}} }}{\text{dx}}}$

माना ${\text{u}}\;{\text{ = }}\;{{\text{x}}^{\text{3}}}$

\begin{align}{\text{du = 3}}{{\text{x}}^{\text{2}}}{\text{dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{3}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + }}{{\left( {{{\text{a}}^{\text{3}}}} \right)}^{\text{2}}}} }}} \;\;\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + b}}{\quad ^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\ \end{align}

\begin{align}{\text{I = }}\dfrac{{\text{1}}}{{\text{3}}}{\text{log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{6}}}} } \right|{\text{ + c \{ y = u, b = }}{{\text{a}}^3}{\text{\} }} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{3}}}{\text{log}}\left| {{{\text{x}}^{\text{3}}}{\text{ + }}\sqrt {{{\text{x}}^{\text{6}}}{\text{ + }}{{\text{a}}^{\text{6}}}} } \right|{\text{ + c}} \hfill \\ \end{align}

9.$\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{\sqrt{{\text{ta}}{{\text{n}}^{\text{2}}}{\text{x + 4}}} }}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{\sqrt{{\text{ta}}{{\text{n}}^{\text{2}}}{\text{x + 4}}} }}{\text{dx}}}$

माना ${\text{u = tanx}}$

\begin{align}{\text{du = se}}{{\text{c}}^{\text{2}}}{\text{x dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + (2}}{{\text{)}}^{\text{2}}}} }}} \;\;\,\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{\text{I = log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + (2}}{{\text{)}}^{\text{2}}}} } \right|{\text{ + c}}\quad {\text{\{ y = u, a = 2\} }} \hfill \\{\text{I = log}}\left| {{\text{tanx + }}\sqrt {{\text{ta}}{{\text{n}}^{\text{2}}}{\text{x + 4}}} {\text{ + c}}} \right| \hfill \\ \end{align}

10. $\dfrac{{\text{1}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 2}}} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 2}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 1 + 1}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + 1}}} }}} \hfill \\ \end{align}

माना ${\text{u}}\;{\text{ = }}\;{\text{x + 1}}$

\begin{align}{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + 1}}} }}} \;\;\;\;\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{\text{I = log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + }}{{\text{1}}^{\text{2}}}} } \right|{\text{ + c}}\quad {\text{\{ y = u, a = 1\} }} \hfill \\{\text{I = log}}\left| {{\text{x + 1 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 2}}} } \right|{\text{ + c}} \hfill \\ \end{align}

11. $\dfrac{{\text{1}}}{{{\text{9}}{{\text{x}}^{\text{2}}}{\text{ + 6x + 5}}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{\text{9}}{{\text{x}}^{\text{2}}}{\text{ + 6x + 5}}}}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{{\text{(3x)}}}^{\text{2}}}{{ + 2 \times 1}} \times {\text{3x + 1 + 4}}}}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{{\text{(3x)}}}^{\text{2}}}{{ + 2 \times 1}} \times {\text{3x + (1}}{{\text{)}}^{\text{2}}}{\text{ + (2}}{{\text{)}}^{\text{2}}}}}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{{\text{(3x + 1)}}}^{\text{2}}}{\text{ + (2}}{{\text{)}}^{\text{2}}}}}} \hfill \\ \end{align}

माना ${\text{u = 3x + 1}}$

\begin{align}{\text{du = 3 dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{3}}}\int {\dfrac{{{\text{du}}}}{{{{\text{u}}^{\text{2}}}{\text{ + (2}}{{\text{)}}^{\text{2}}}}}} \hfill \\ \end{align}

\begin{align}\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{u}}}{{\text{2}}}} \right){\text{ + c}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3x + 1}}}}{{\text{2}}}} \right){\text{ + c}} \hfill \\ \end{align}

12. $\dfrac{{\text{1}}}{{\sqrt {{\text{7 - 6x - }}{{\text{x}}^{\text{2}}}} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{7 - 6x - }}{{\text{x}}^{\text{2}}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{16 - 9 - 6x - }}{{\text{x}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{{\text{(4)}}}^{\text{2}}}{\text{ - (9}}{{\text{)}}^{\text{2}}}{{ - 2 \times 3 \times x - }}{{\text{x}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{{\text{(4)}}}^{\text{2}}}{\text{ - (3 + x}}{{\text{)}}^{\text{2}}}} }}} \hfill \\ \end{align}

माना ${\text{u = 3 + x}}$

\begin{align}{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{du}}}}{{{{\sqrt {{\text{(4)}}} }^{\text{2}}}{\text{ - }}{{\text{u}}^{\text{2}}}}}} \;\,\;\,\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - y}}} }^{\text{2}}}}}} {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{u}}}{{\text{4}}}} \right){\text{ + c}} \hfill \\ \end{align}

${\text{I = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{3 + {\text{x}}}}{{\text{4}}}} \right){\text{ + c}}$

13. $\dfrac{{\text{1}}}{{\sqrt {{\text{(x - 1)(x - 2)}}} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{(x - 1)(x - 2)}}} }}}$

${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 3x + 2}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ - 2 \times }}\left( {\dfrac{{\text{3}}}{{\text{2}}}} \right){{ \times x + }}{{\left( {\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ + 2}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ + 2}}} }}} {\text{ = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{4}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \hfill \\ \end{align}

माना ${\text{u = x - }}\dfrac{{\text{3}}}{{\text{2}}}$

\begin{align}{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{{\text{(u)}}}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \;\;\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^2}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\ \end{align}

\begin{align}{\text{I = log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right){\text{2}}} } \right|{\text{ + c \{ y = u, a = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{\} }} \hfill \\{\text{I = log}}\left| {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}{\text{ + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 3x + 2}}} } \right|{\text{ + c}} \hfill \\ \end{align}

14. $\dfrac{{\text{1}}}{{\sqrt {{\text{8 + 3x - x}}} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{8 + 3x - }}{{\text{x}}^{\text{2}}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{8 + }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ + 3x - }}{{\text{x}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {\dfrac{{{\text{41}}}}{{\text{4}}}{\text{ - }}{{\left( {\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ + 3x - }}{{\text{x}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {\dfrac{{\sqrt {{\text{41}}} }}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \hfill \\ \end{align}

माना ${\text{u = x - }}\dfrac{{\text{3}}}{{\text{2}}}$

\begin{align}{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{\left( {\dfrac{{\sqrt {{\text{41}}} }}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\text{u}}^{\text{2}}}} }}} \;\;\;\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}} {\text{ = si}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = si}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{u}}}{{\dfrac{{\sqrt {{\text{41}}} }}{{\text{2}}}}}} \right){\text{ + c \{ y = u, a = }}\dfrac{{\sqrt {{\text{41}}} }}{{\text{2}}}{\text{\} }} \hfill \\ \end{align}

\begin{align}{\text{I = sin - 1}}\left( {\dfrac{{{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\sqrt {{\text{41}}} }}{{\text{2}}}}}} \right){\text{ + c}} \hfill \\{\text{I = si}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{2x - 3}}}}{{\sqrt {{\text{41}}} }}} \right){\text{ + c}} \hfill \\ \end{align}

15. $\dfrac{{\text{1}}}{{\sqrt {{\text{(x - a)(x - b)}}} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{(x - a)(x - b)}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - (a + b)x + ab}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ - 2 \times }}\left( {\dfrac{{{\text{a + b}}}}{{\text{2}}}} \right){{ \times x + }}{{\left( {\dfrac{{{\text{a + b}}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{{\text{a + b}}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ + ab}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{{\text{a + b}}}}{{\text{2}}}} \right)}^2}{\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}}}{{\text{4}}}{\text{ + ab}}} }}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}{\text{ - 2ab + }}{{\text{b}}^{\text{2}}}}}{{\text{4}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{{\text{a + b}}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{{\text{a - b}}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \hfill \\ \end{align}

माना ${\text{u}}\;{\text{ = }}\;{\text{x - }}\dfrac{{{\text{a + b}}}}{{\text{2}}}$

\begin{align}{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{{\text{(u)}}}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{{\text{a - b}}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \;\;\;\;\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{d}}^{\text{2}}}} }}} {\text{ = log|y + }}\sqrt {{{\text{y}}^2}{\text{ - }}{{\text{d}}^2}} {\text{| + c}}} \right\} \hfill \\{\text{I = log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^2}{\text{ - }}{{\left( {\dfrac{{{\text{a - b}}}}{{\text{2}}}} \right)}^{\text{2}}}} } \right|{\text{ + c}}\;\,{\text{ }}\;{\text{\{ y = u, d = }}\dfrac{{{\text{a - b}}}}{{\text{2}}}{\text{\} }} \hfill \\{\text{I = log}}\left| {{\text{x - }}\dfrac{{{\text{a + b}}}}{{\text{2}}}{\text{ + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - (a + b)x + ab}}} } \right|{\text{ + c}} \hfill \\{\text{I = log}}\left| {{\text{x - }}\dfrac{{{\text{a + b}}}}{{\text{2}}}{\text{ + }}\sqrt {{\text{(x - a)(x - b)}}} } \right|{\text{ + c}} \hfill \\ \end{align}

16. $\dfrac{{{\text{4x + 1}}}}{{\sqrt {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + x - 3}}} }}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{4x + 1}}}}{{\sqrt {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + x - 3}}} }}} {\text{dx}}$

माना ${\text{u}}\;{\text{ = }}\;{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + x - 3}}$

\begin{align}{\text{du}}\;{\text{ = }}\;{\text{(4x + 1)dx}} \hfill \\{\text{I}}\;{\text{ = }}\;\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} \hfill \\ \end{align}

\begin{align}{\text{I}}\;{\text{ = }}\;{\text{2}}\sqrt {\text{u}} {\text{ + c}} \hfill \\{\text{I}}\;{\text{ = }}\;{\text{2}}\sqrt {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ + x - 3}}} {\text{ + c}} \hfill \\ \end{align}

17. $\dfrac{{{\text{x + 2}}}}{{\sqrt {{{\text{x}}^2} - 1} }}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(x + 2)}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{dx}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{xdx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{ + }}\int {\dfrac{{{\text{2dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{xdx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} {\text{ , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{2dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} \hfill \\{\text{I = }}{{\text{I}}_1}{\text{ + }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{xdx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}} \hfill \\ \end{align}

माना ${\text{u = }}{{\text{x}}^2}{\text{ - 1}}$

${\text{du = 2x dx}}$

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times (2}}\sqrt {\text{u}} {\text{) + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\sqrt {\text{u}} {\text{ + }}{{\text{c}}_1} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{2dx}}}}{{\sqrt {{{\text{x}}^2} - 1} }}} \;\;\,\,\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{d}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{d}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}\;{\text{ = }}\;{\text{2log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} } \right|{\text{ + }}{{\text{c}}_2}\quad {\text{\{ y = x, d = 1\} }} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}} \hfill \\{\text{I = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} {\text{ + }}{{\text{c}}_1}{\text{ + 2log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} {\text{ + 2log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} } \right|{\text{ + C}} \hfill \\ \end{align}

18. $\dfrac{{{\text{5x - 2}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(5x - 2)dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}}$

\begin{align}{\text{I = }}\int {\dfrac{{\left[ {\dfrac{{\text{5}}}{{\text{6}}}{\text{(6x + 2 - 2) - 2}}} \right]{{ \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} \hfill \\{\text{I = }}\int {\dfrac{{\left[ {\dfrac{{\text{5}}}{{\text{6}}}{\text{(6x + 2) - }}\dfrac{{{\text{10}}}}{{\text{6}}}{\text{ - 2}}} \right]{{ \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} \hfill \\{\text{I = }}\int {\dfrac{{\left[ {\dfrac{{\text{5}}}{{\text{6}}}{\text{(6x + 2) - }}\dfrac{{{\text{11}}}}{{\text{3}}}} \right]{{ \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} \hfill \\{\text{I = }}\dfrac{{\text{5}}}{{\text{6}}}\int {\dfrac{{{{(6x + 2) \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} {\text{ - }}\dfrac{{{\text{11}}}}{{\text{3}}}\int {\dfrac{{{\text{dx}}}}{{{\text{1 + 2x + 3x}}}}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{6}}}\int {\dfrac{{{{(6x + 2) \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} \hfill \\{{\text{I}}_2}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{3}}}\int {\dfrac{{{\text{dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}} \hfill \\ \end{align}

${\text{I = }}{{\text{I}}_1}{\text{ - }}{{\text{I}}_2}$

${{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{6}}}\int {\dfrac{{{{(6x + 2) \times dx}}}}{{{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}}}}$

${\text{u = 1 + 2x + 3}}{{\text{x}}^2}$

\begin{align}{\text{du = (6x + 2)dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{6}}}\int {\dfrac{{{\text{du}}}}{{\text{u}}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{6}}}{\text{log|u| + c}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{6}}}{\text{log}}\left| {{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}} \right|{\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{3}}}\int {\dfrac{{{\text{dx}}}}{{{\text{3}}{{\text{x}}^{\text{2}}}{\text{ + 2x + 1}}}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{3}}}\int {\dfrac{{{\text{dx}}}}{{{\text{3}}\left( {{{\text{x}}^{\text{2}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}}{\text{x + }}\dfrac{{\text{1}}}{{\text{3}}}} \right)}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{9}}}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{{ + 2 \times }}\dfrac{{\text{1}}}{{\text{3}}}{{ \times x + }}\dfrac{{\text{1}}}{{\text{9}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{9}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{3}}}}}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{9}}}\int {\dfrac{{{\text{dx}}}}{{{{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{3}}}} \right)}^{\text{2}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{9}}}}}} \hfill \\{{\text{I}}_2}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{9}}}\int {\dfrac{{{\text{dx}}}}{{{{\left( {{\text{x + }}\dfrac{{\text{1}}}{{\text{3}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}} \right)}^{\text{2}}}}}} \hfill \\{\text{u = x + }}\dfrac{{\text{1}}}{{\text{3}}} \hfill \\{\text{du = dx}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{9}}}\int {\dfrac{{{\text{dx}}}}{{{{\text{u}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}} \right)}^{\text{2}}}}}} \;\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{\text{9}}}{{ \times }}\dfrac{{\text{1}}}{{\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}}}{{ \times ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{u}}}{{\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}}}} \right){\text{ + }}{{\text{c}}_2}\quad {\text{\{ y = u, a = }}\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}{\text{\} }} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{3}}\sqrt {\text{2}} }}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x + }}\dfrac{{\text{1}}}{{\text{3}}}}}{{\dfrac{{\sqrt {\text{2}} }}{{\text{3}}}}}} \right){\text{ + c}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{11}}}}{{{\text{3}}\sqrt {\text{2}} }}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3x + 1}}}}{{\sqrt {\text{2}} }}} \right){\text{ + c}} \hfill \\ \end{align}

\begin{align}{\text{I = }}{{\text{I}}_1}{\text{ - }}{{\text{I}}_2} \hfill \\{\text{I = }}\dfrac{{\text{5}}}{{\text{6}}}{\text{log|1 + 2x + 3}}{{\text{x}}^2}{\text{| + }}{{\text{c}}_1}{\text{ - }}\dfrac{{{\text{11}}}}{{{\text{3}}\sqrt {\text{2}} }}{{ \times ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{3x + 1}}}}{{\sqrt {\text{2}} }}} \right){\text{ - }}{{\text{c}}_2} \hfill \\ \end{align}

${\text{I = }}\dfrac{{\text{5}}}{{\text{6}}}{\text{log}}\left| {{\text{1 + 2x + 3}}{{\text{x}}^{\text{2}}}} \right|{\text{ - }}\dfrac{{{\text{11}}}}{{{\text{3}}\sqrt {\text{2}} }}{{ \times ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{3x + 1}}}}{{\sqrt {\text{2}} }}} \right){\text{ + C}}$

19. $\mathbf{\dfrac{{{\text{6x + 7}}}}{{\sqrt {{\text{(x - 5)(x - 4)}}} }}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(6x + 7)dx}}}}{{\sqrt {{\text{(x - 5)(x - 4)}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{(6x + 7)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(6x - 27 + 27 + 7)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(6x - 27)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} {\text{ + }}\int {\dfrac{{{\text{34}}\;{\text{dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(6x - 27)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{34 dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(6x - 27)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{3(2x - 9)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\{\text{u = }}{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}} \hfill \\{\text{du = (2x - 9)dx}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{3du}}}}{{\sqrt {\text{u}} }}} {\text{ - }}{{\text{I}}_{\text{1}}}{\text{ = 3 \times 2}}\sqrt {\text{u}} {\text{ + }}{{\text{c}}_{\text{1}}}{{\text{I}}_{\text{1}}}{\text{ = 6}}\sqrt {{{\text{x}}^2}{\text{ - 9x + 20}}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{34 dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} }}} \hfill \\\end{align}

${{\text{I}}_2}{\text{ = }}\int {\dfrac{{{\text{34 dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 2 \times }}\dfrac{{\text{9}}}{{\text{2}}}{\text{ \times x + }}{{\left( {\dfrac{{\text{9}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{9}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ + 20}}} }}}$

${{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{34 dx}}}}{{\sqrt {{{\left( {{\text{x - }}\dfrac{{\text{9}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{4}}}} }}}$

\begin{align}{\text{u = x - }}\dfrac{{\text{9}}}{{\text{2}}} \hfill \\{\text{du = dx}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{34 dx}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}} }}} \;\,\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{d}}^{\text{2}}}} }}} {\text{ = log|y + }}\sqrt {{{\text{y}}^2}{\text{ - }}{{\text{d}}^2}} {\text{| + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 34log}}\left| {{\text{u + }}\sqrt {{\text{u}}\quad {\text{2 - }}{{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\ \end{align}

\begin{align}{\text{\{ y = u, d = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{\} }} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 34log}}\left| {{\text{x - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{ + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9x + 20}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\{\text{I = 6}}\sqrt {{{\text{x}}^2}{\text{ - 9x + 20}}} {\text{ + c + 34log}}\left| {{\text{x - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{ + }}\sqrt {{{\text{x}}^2}{\text{ - 9x + 20}}} } \right|{\text{ + }}{{\text{c}}_2} \hfill \\{\text{I = 6}}\sqrt {{{\text{x}}^2}{\text{ - 9x + 20}}} {\text{ + 34log}}\left| {{\text{x - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{ + }}\sqrt {{{\text{x}}^2}{\text{ - 9x + 20}}} } \right|{\text{ + C}} \hfill \\ \end{align}

20. $\mathbf{\dfrac{{{\text{x + 2}}}}{{\sqrt {{\text{4x - }}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{\text{4x - x}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{\text{4 - 4 + 4x - }}{{\text{x}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (2}}{{\text{)}}^{\text{2}}}{{ + 2 \times 2 \times x - (x}}{{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} \hfill \\ \end{align}

${\text{I = }}\int {\dfrac{{{\text{(x - 2 + 4)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}}$

${\text{I = }}\int {\dfrac{{{\text{(x - 2)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} {\text{ + }}\int {\dfrac{{{\text{4dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}}$

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x - 2)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} {\text{ , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{4dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x - 2)dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{u = (2}}{{\text{)}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}} \hfill \\{\text{du = - 2(x - 2)dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}{{ \times 2 \times }}\sqrt {\text{u}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = - }}\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} {\text{ + c}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = - }}\sqrt {{\text{4x - }}{{\text{x}}^{\text{2}}}} {\text{ + c}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{4dx}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x - 2}}{{\text{)}}^{\text{2}}}} }}} \hfill \\ {\text{u = x - 2}} \hfill \\ {\text{du = dx}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 4}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - }}{{\text{u}}^{\text{2}}}} }}} \;\;\,\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}} {\text{ = si}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 4si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{u}}}{{\text{2}}}} \right){\text{ + }}{{\text{c}}_2}\quad {\text{\{ y = u, a = 2\} }} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 4si}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{\text{x - 2}}}}{{\text{2}}}} \right){\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}} \hfill \\{\text{I = - }}\sqrt {{\text{4x - }}{{\text{x}}^{\text{2}}}} {\text{ + }}{{\text{c}}_{\text{1}}}{\text{ + 4si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x - 2}}}}{{\text{2}}}} \right){\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = - }}\sqrt {{\text{4x - x}}\quad {\text{2}}} {\text{ + 4si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x - 2}}}}{{\text{2}}}} \right){\text{ + C}} \hfill \\ \end{align}

21. $\mathbf{\dfrac{{{\text{x + 2}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} }}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} }}}$

${\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ + 2 \times x \times 1 + 1 + 2}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ + 2 \times x \times 1 + (1}}{{\text{)}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x + 2)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x + 1)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \; + \;\int {\dfrac{{{\text{ (1) dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x + 1)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} {\text{ , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{(1)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x + 1)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{u = (x + 1}}{{\text{)}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}} \hfill \\ \end{align}

\begin{align}{\text{du = 2(x + 1)dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times 2 \times }}\sqrt {\text{u}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} {\text{ + }}{{\text{c}}_{\text{1}}}\;,\;{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{(1)dx}}}}{{\sqrt {{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{u = (x + 1)}} \hfill \\{\text{du = dx}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} }}} \;\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}}\;\;\;{\text{\{ y = u, a = }}\sqrt {\text{2}} {\text{\} }} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = log}}\left| {{\text{x + 1 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\ \end{align}

${\text{I}}\;{\text{ = }}\;{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}$

\begin{align}{\text{I = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} {\text{ + }}{{\text{c}}_{\text{1}}}{\text{ + log}}\left| {{\text{x + 1 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} {\text{ + log}}\left| {{\text{x + 1 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 2x + 3}}} } \right|{\text{ + C}} \hfill \\ \end{align}

22. $\mathbf{\dfrac{{{\text{x + 3}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 2x - 5}}}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(x + 3)dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 2x - 5}}}}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{(x + 3)dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 2 \times 1 \times x + (1}}{{\text{)}}^{\text{2}}}{\text{ - (1}}{{\text{)}}^{\text{2}}}{\text{ - 5}}}}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x - 1 + 1 + 3)dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(x - 1)dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} {\text{ + }}\int {\dfrac{{{\text{4dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x - 1)dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} {\text{ , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{4dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(x - 1)dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \hfill \\ \end{align}

\begin{align}{\text{u = (x - 1}}{{\text{)}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}} \hfill \\{\text{du = 2(x - 1)}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\text{u}}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log|u| + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ - 2x - 5}}} \right|{\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{4dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \hfill \\{\text{u = (x - 1)}} \hfill \\{\text{du = dx}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 4}}\int {\dfrac{{{\text{du}}}}{{{{\text{u}}^{\text{2}}}{\text{- (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}}}} \quad \left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{{\text{2a}}}}{\text{log}}\left| {\dfrac{{{\text{x - a}}}}{{{\text{x + a}}}}} \right|{\text{ + c}}} \right\} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 4 \times }}\dfrac{{\text{1}}}{{{\text{2}}\sqrt {\text{6}} }}{\text{log}}\left| {\dfrac{{{\text{u - }}\sqrt {\text{6}} }}{{{\text{u + }}\sqrt {\text{6}} }}} \right|{\text{ + }}{{\text{c}}_{\text{2}}}\quad {\text{\{ y = u, a = }}\sqrt {\text{6}} {\text{\} }} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{\text{2}}}{{\sqrt {\text{6}} }}{\text{log}}\left| {\dfrac{{{\text{x - 1 - }}\sqrt {\text{6}} }}{{{\text{x - 1 + }}\sqrt {\text{6}} }}} \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_2} \hfill \\ \end{align}

${\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ - 2x - 5}}} \right|{\text{ + }}{{\text{c}}_{\text{1}}}{\text{ + }}\dfrac{{\text{2}}}{{\sqrt {\text{6}} }}{\text{log}}\left| {\dfrac{{{\text{x - 1 - }}\sqrt {\text{6}} }}{{{\text{x - 1 + }}\sqrt {\text{6}} }}} \right|{\text{ + }}{{\text{c}}_{\text{2}}}$

${\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ - 2x - 5}}} \right|{\text{ + }}\dfrac{{\text{2}}}{{\sqrt {\text{6}} }}{\text{log}}\left| {\dfrac{{{\text{x - 1 - }}\sqrt {\text{6}} }}{{{\text{x - 1 + }}\sqrt {\text{6}} }}} \right|{\text{ + C}}$

23. $\mathbf{\dfrac{{{\text{5x + 3}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} }}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{(5x + 3)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{(5x + 3)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ + 2 \times 2 \times x + 4 + 6}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(5x + 3)dx}}}}{{\sqrt {{{\text{x}}^{\text{2}}}{{ + 2 \times 2 \times x + (2}}{{\text{)}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(5x + 3)dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(5x + 10 - 10 + 3)dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{(5x + 10)dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} {\text{ - }}\int {\dfrac{{{\text{7dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(5x + 10)dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} {\text{ , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{7dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ - }}{{\text{I}}_2} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{(5x + 10)dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{u = (x + 2}}{{\text{)}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}} \hfill \\{\text{du = 2(x + 2)dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {\text{u}} }}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{5}}}{{\text{2}}}{{ \times 2 \times }}\sqrt {\text{u}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = 5}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} {\text{ + }}{{\text{c}}_{\text{1}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int {\dfrac{{{\text{7dx}}}}{{\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{u = (x + 2)}} \hfill \\{\text{du = dx}} \hfill \\ \end{align}

${{\text{I}}_{\text{2}}}{\text{ = 7}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{\text{u}}^{\text{2}}}{\text{ + (}}\sqrt {\text{6}} {{\text{)}}^{\text{2}}}} }}} \;\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} }}} {\text{ = log}}\left| {{\text{y + }}\sqrt {{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + c}}} \right\}$

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = 7log}}\left| {{\text{u + }}\sqrt {{{\text{u}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = 7log}}\left| {{\text{x + 2 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} } \right|{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ - }}{{\text{I}}_{\text{2}}} \hfill \\\begin{array}{*{20}{l}}{{\text{I = 5}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} {\text{ + }}{{\text{c}}_{\text{1}}}{\text{ - 7log}}\mid {\text{x + 2 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} } \end{array}{\text{ + }}{{\text{c}}_{\text{2}}} \hfill \\{\text{I = 5}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} {\text{ - 7log}}\left| {{\text{x + 2 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 10}}} } \right|{\text{ + C}} \hfill \\ \end{align}

24 एवं 25 मे सही उत्तर का चयन कीजिए:

24. $\mathbf{\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 2x + 2}}}}} }$ बराबर है।

(a) $\mathbf{{\text{xta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x + 1) + c}}}$

(b) $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x + 1) + c}}}$

(c) $\mathbf{{\text{(x + 1)ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + c}}}$

(d) $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + c}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$$\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 2x + 2}}}}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{{ + 2 \times 1 \times x + 1 + 1}}}}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{{ + 2 \times 1 \times x + (1}}{{\text{)}}^{\text{2}}}{\text{ + (1}}{{\text{)}}^{\text{2}}}}}} \hfill \\ \end{align}

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + (1}}{{\text{)}}^{\text{2}}}}}} \hfill \\{\text{u = (x + 1)}} \hfill \\{\text{du = dx}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{du}}}}{{{{\text{u}}^{\text{2}}}{\text{ + (1}}{{\text{)}}^{\text{2}}}}}} \;\;\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ tan}}{{\text{ }}^{ - 1}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{u}}}{{\text{1}}}} \right){\text{ + c}}\quad {\text{\{ y = u, a = 1\} }} \hfill \\ \end{align}

${\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x + 1) + c}}$

अतः विकल्प (b) सही है।

25. $\mathbf{\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{9x - 4}}{{\text{x}}^{\text{2}}}} }}} }$ बराबर है।

1. $\mathbf{\dfrac{{\text{1}}}{{\text{9}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{9x - 8}}}}{{\text{8}}}} \right){\text{ + c}}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{8x - 9}}}}{{\text{9}}}} \right){\text{ + c}}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\text{3}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{9x - 8}}}}{{\text{8}}}} \right){\text{ + c}}}$

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{9x - 8}}}}{{\text{9}}}} \right){\text{ + c}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{9x - 4}}{{\text{x}}^{\text{2}}}} }}}$

\begin{align}{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {\dfrac{{\text{9}}}{{\text{4}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{9}}}{{\text{4}}}} \right)}^{\text{2}}}{{ + 2 \times }}\dfrac{{\text{9}}}{{\text{4}}}{{ \times 2x - (2x}}{{\text{)}}^{\text{2}}}} }}} \hfill \\{\text{I = }}\int {\dfrac{{{\text{dx}}}}{{\sqrt {{{\left( {\dfrac{{\text{9}}}{{\text{4}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {{\text{2x - }}\dfrac{{\text{9}}}{{\text{4}}}} \right)}^{\text{2}}}} }}} \hfill \\ \end{align}

\begin{align}{\text{u = }}\left( {{\text{2x - }}\dfrac{{\text{9}}}{{\text{4}}}} \right) \hfill \\{\text{du = 2dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{du}}}}{{\sqrt {{{\left( {\dfrac{{\text{9}}}{{\text{4}}}} \right)}^{\text{2}}}{\text{ - }}{{\text{u}}^{\text{2}}}} }}} \;\;\,\;\;\left\{ {\int {\dfrac{{{\text{dy}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}} {\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{y}}}{{\text{a}}}} \right){\text{ + c}}} \right\} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{u}}}{{\dfrac{{\text{9}}}{{\text{4}}}}}} \right){\text{ + c \{ y = u, a = }}\dfrac{{\text{9}}}{{\text{4}}}{\text{\} }} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x - }}\dfrac{{\text{9}}}{{\text{4}}}}}{{\dfrac{{\text{9}}}{{\text{4}}}}}} \right){\text{ + c}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{8x - 9}}}}{{\text{9}}}} \right){\text{ + c}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

### 1 से 21 तक के प्रश्नों मे परीमी फलनों का समाकलन कीजिए:

1. $\mathbf{\dfrac{{\text{x}}}{{{\text{(x + 1)(x + 2)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{x}}}{{{\text{(x + 1)(x + 2)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x + 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x + 2)}}}} \hfill \\\dfrac{{\text{x}}}{{{\text{(x + 1)(x + 2)}}}}{\text{ = }}\dfrac{{{\text{A(x + 2)}}}}{{{\text{(x + 1)}}}}{\text{ + }}\dfrac{{{\text{B(x + 1)}}}}{{{\text{(x + 2)}}}} \hfill \\{\text{x = A(x + 2) + B(x + 1)}} \hfill \\ \end{align}

\begin{align}{\text{x = - 2 }} \Rightarrow {\text{ - 2 = B( - 2 + 1) }} \Rightarrow {\text{ B = 2}} \hfill \\{\text{x = - 1 }} \Rightarrow {\text{ - 1 = A( - 1 + 2) }} \Rightarrow \;{\text{A = - 1}} \hfill \\\dfrac{{\text{x}}}{{{\text{(x + 1)(x + 2)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{(x + 1)}}}}{\text{ + }}\dfrac{{\text{2}}}{{{\text{(x + 2)}}}} \hfill \\\int {\dfrac{{{\text{xdx}}}}{{{\text{(x + 1)(x + 2)}}}}} {\text{ = - }}\int {\dfrac{{{\text{1dx}}}}{{{\text{(x + 1)}}}}} {\text{ + 2}}\int {\dfrac{{{\text{dx}}}}{{{\text{(x + 2)}}}}} \hfill \\{\text{ = - log(x + 1) + 2log(x + 2) + c}} \hfill \\ \end{align}

2. $\mathbf{\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ - 9}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ - 9}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{x + 3}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{x - 3}}}} \hfill \\{\text{1 = A(x - 3) + B(x + 3)}} \hfill \\{\text{x = - 3 }} \Rightarrow {\text{ 1 = A( - 3 - 3) }} \Rightarrow {\text{ A = - 1/6}} \hfill \\{\text{x = 3 }} \Rightarrow {\text{ 1 = B(3 + 3) }} \Rightarrow \;{\text{B = 1/6}} \hfill \\ \end{align}

\begin{align}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ - 9}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{6(x + 3)}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{6(x - 3)}}}} \hfill \\\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 9}}}}} {\text{ = - }}\int {\dfrac{{{\text{dx}}}}{{{\text{6(x + 3)}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{6(x + 2)}}}}} \hfill \\ \end{align}

$\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 9}}}}} {\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{6}}}{\text{log(x + 3) + }}\dfrac{{\text{1}}}{{\text{6}}}{\text{log(x - 3) + c}}$

${\text{ = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{log}}\left( {\dfrac{{{\text{x - 3}}}}{{{\text{x + 3}}}}} \right){\text{ + c}}$

3. $\mathbf{\dfrac{{{\text{3x - 1}}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{3x - 1}}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x - 2)}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{(x - 3)}}}} \hfill \\{\text{3x - 1 = A(x - 3)(x - 2) + B(x - 1)(x - 3) + C(x - 1)(x - 2)}} \hfill \\{\text{x = 1 }} \Rightarrow {\text{ 2 = A( - 2)( - 1) }} \Rightarrow \;{\text{A = 1}} \hfill \\{\text{x = 2 }} \Rightarrow \;{\text{5 = B( - 1)(1) }} \Rightarrow \;{\text{B = - 5}} \hfill \\{\text{x = 3 }} \Rightarrow \;{\text{8 = C(2)(1) }} \Rightarrow \;{\text{C = 4}} \hfill \\\dfrac{{{\text{3x - 1}}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{{\text{ - 5}}}}{{{\text{(x - 2)}}}}{\text{ + }}\dfrac{{\text{4}}}{{{\text{(x - 3)}}}} \hfill \\ \end{align}

\begin{align}\int {\dfrac{{{\text{(3x - 1)dx}}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}} {\text{ = }}\int {\dfrac{{{\text{1dx}}}}{{{\text{(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{ - 5dx}}}}{{{\text{(x - 2)}}}}} {\text{ + }}\int {\dfrac{{{\text{4dx}}}}{{{\text{(x - 3)}}}}} \hfill \\{\text{ = log(x - 1) - 5log(x - 2) + 4log(x - 3)}} \hfill \\{\text{ = log(}}\dfrac{{{\text{(x - 1)(x - 3}}{{\text{)}}^{\text{4}}}}}{{{{{\text{(x - 2)}}}^{\text{5}}}}}{\text{) + c}} \hfill \\ \end{align}

4. $\mathbf{\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x - 2)}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{(x - 3)}}}} \hfill \\{\text{x = A(x - 3)(x - 2) + B(x - 1)(x - 3) + C(x - 1)(x - 2)}} \hfill \\{\text{x = 1 }} \Rightarrow {\text{ 1 = A( - 2)( - 1) }} \Rightarrow \;{\text{A = 1/2}} \hfill \\{\text{x = 2 }} \Rightarrow \;{\text{2 = B( - 1)(1) }} \Rightarrow \;{\text{B = - 2}} \hfill \\{\text{x = 3 }} \Rightarrow \;{\text{3 = C(2)(1) }} \Rightarrow \;{\text{C = 3/2}} \hfill \\ \end{align}

$\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{2(x - 1)}}}}{\text{ + }}\dfrac{{{\text{ - 2}}}}{{{\text{(x - 2)}}}}{\text{ + }}\dfrac{{\text{3}}}{{{\text{2(x - 3)}}}}$

$\int {\dfrac{{{\text{xdx}}}}{{{\text{(x - 1)(x - 2)(x - 3)}}}}} {\text{ = }}\int {\dfrac{{{\text{dx}}}}{{{\text{2(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{ - 2dx}}}}{{{\text{(x - 2)}}}}} {\text{ + }}\int {\dfrac{{{\text{3dx}}}}{{{\text{2(x - 3)}}}}}$

${\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(x - 1) - 2log(x - 2) + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{log(x - 3) + C}}$

5. $\mathbf{\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 3x + 2}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

6. $\mathbf{\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{x(1 - 2x)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{x(1 - 2x)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{x(1 - 2x)}}}}{\text{ + }}\dfrac{{\text{x}}}{{{\text{(1 - 2x)}}}} \hfill \\\dfrac{{\text{1}}}{{{\text{x(1 - 2x)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(1 - 2x)}}}} \hfill \\{\text{x = 0 }} \Rightarrow \;{\text{1 = A(1 - 2x) + Bx}} \hfill \\ \end{align}

\begin{align}{\text{1 = A}} \hfill \\{\text{x = 1/2 }} \Rightarrow {\text{ 1 = B(1/2) }} \Rightarrow {\text{ B = 2}} \hfill \\\dfrac{{\text{x}}}{{{\text{(1 - 2x)}}}}{\text{ = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2(1 - 2x)}}}} \hfill \\ \end{align}

${\text{ = }}\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{x(1 - 2x)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{x(1 - 2x)}}}}{\text{ + }}\dfrac{{\text{x}}}{{{\text{(1 - 2x)}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{2}}}{{{\text{(1 - 2x)}}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2(1 - 2x)}}}}$

$\int {\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{{\text{x(1 - 2x)}}}}} {\text{ = }}\int {\dfrac{{{\text{2dx}}}}{{{\text{(1 - 2x)}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{2(1 - 2x)}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{\text{x}}}} {\text{ - }}\int {\dfrac{{{\text{dx}}}}{{\text{2}}}}$

\begin{align}{\text{ = log(1 - 2x) - }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log(1 - 2x) + log(x) - }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + c}} \hfill \\{\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}{\text{log(1 - 2x) + log(x) - }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + c}} \hfill \\ \end{align}

7. $\mathbf{\dfrac{{\text{x}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(x - 1)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{x}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(x - 1)}}}}{\text{ = }}\dfrac{{{\text{Ax + C}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x - 1)}}}} \hfill \\{\text{x = (Ax + C)(x - 1) + B}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right) \hfill \\{\text{x = 1 }} \Rightarrow \;1 = 2{\text{B }} \Rightarrow {\text{ B = 1/2}} \hfill \\{\text{x = 0 }} \Rightarrow \;0 = \;{\text{C}} - {\text{B}}\; \Rightarrow \;1/2 \hfill \\{\text{x = - 1 }} \Rightarrow \;\dfrac{{ - 1}}{4}\; = \;\dfrac{{{\text{ - (A + C)}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{B}}}{{\text{2}}}\; \Rightarrow \; - {\text{A + C - B}}\; = \; - 1/2\; \Rightarrow \;{\text{A}}\; = \;1/2\;\; \hfill \\ \end{align}

\begin{align}\dfrac{{\text{x}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(x - 1)}}}}{\text{ = }}\dfrac{{{\text{( - x + 1)}}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2(x - 1)}}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - x}}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2(x - 1)}}}} \hfill \\\int {\dfrac{{{\text{xdx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(x - 1)}}}}} {\text{ = }}\int {\dfrac{{{\text{ - xdx}}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} \int {\dfrac{{{\text{dx}}}}{{{\text{2(x - 1)}}}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{4}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(x - 1) + C}} \hfill \\ \end{align}

8. $\mathbf{\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 1)(x + 2)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

$\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 1)(x + 2)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{(x + 2)}}}}$

\begin{align}{\text{x = A(x + 2)(x - 1) + B(x + 2) + C(x - 1}}{{\text{)}}^{\text{2}}} \hfill \\{\text{x = 1 }} \Rightarrow \;{\text{1 = A(3)(0) + B(3) + 0 }} \Rightarrow {\text{ B = 1/3}} \hfill \\{\text{x = - 2 }} \Rightarrow {\text{ - 2 = C(9) }} \Rightarrow {\text{ C = - 2/9}} \hfill \\ \end{align}

\begin{align}{\text{x = 0 }} \Rightarrow {\text{ 0 = A(2)( - 1) + B(2) + C }} \Rightarrow {\text{ - 2A = C - 2B = - 4/9 }} \Rightarrow {\text{ A = 2/9}} \hfill \\\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 1)(x + 2)}}}}{\text{ = }}\dfrac{{\text{2}}}{{{\text{9(x - 1)}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{3(x - 1}}{{\text{)}}^{\text{2}}}}}{\text{ + }}\dfrac{{{\text{ - 2}}}}{{{\text{9(x + 2)}}}} \hfill \\\int {\dfrac{{\text{x}}}{{{\text{(x - 1)(x - 1)(x + 2)}}}}} {\text{ = }}\int {\dfrac{{{\text{2dx}}}}{{{\text{9(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{1dx}}}}{{{\text{3(x - 1}}{{\text{)}}^{\text{2}}}}}} {\text{ + }}\int {\dfrac{{{\text{ - 2}}}}{{{\text{9(x + 2)}}}}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{9}}}{\text{log(x - 1) - }}\dfrac{{\text{1}}}{{{\text{3(x - 1)}}}}{\text{ + }}\dfrac{{{\text{ - 2}}}}{{\text{9}}}{\text{log(x + 2) + c}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{9}}}{\text{log}}\dfrac{{{\text{(x - 1)}}}}{{{\text{(x + 2)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{3(x - 1)}}}}{\text{ + c}} \hfill \\ \end{align}

9. $\mathbf{\dfrac{{{\text{3x + 5}}}}{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - x + 1}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{3x + 5}}}}{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - x + 1}}}}{\text{ = }}\dfrac{{{\text{3x + 5}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}{\text{(x + 1)}}}} \hfill \\\dfrac{{{\text{3x + 5}}}}{{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - x + 1}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{(x + 1)}}}} \hfill \\ \end{align}

\begin{align}{\text{3x + 5 = A(x + 1)(x - 1) + B(x + 1) + C(x - 1}}{{\text{)}}^{\text{2}}} \hfill \\{\text{x = 1 }} \Rightarrow {\text{ 8 = B(2) }} \Rightarrow {\text{ B = 4}} \hfill \\{\text{x = - 1 }} \Rightarrow {\text{ 2 = C(4) }} \Rightarrow {\text{ C = 1/2}} \hfill \\{\text{x = 0 }} \Rightarrow {\text{ 5 = - A + B + C }} \Rightarrow {\text{ A = - 1/2}} \hfill \\\dfrac{{{\text{3x + 5}}}}{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - x + 1}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{2(x - 1)}}}}{\text{ + }}\dfrac{{\text{4}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{2(x + 1)}}}} \hfill \\\int {\dfrac{{{\text{(3x + 5)dx}}}}{{{{\text{x}}^{\text{3}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ - x + 1}}}}} {\text{ = }}\int {\dfrac{{{\text{ - 1dx}}}}{{{\text{2(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{4dx}}}}{{{{{\text{(x - 1)}}}^{\text{2}}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{2(x + 2)}}}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}{\text{log(x - 1) + 4}}\dfrac{{{\text{( - 1)}}}}{{{\text{x - 1}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(x + 2) + c}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{{\text{(x + 2)}}}}{{{\text{(x - 1)}}}}{\text{ - }}\dfrac{{\text{4}}}{{{\text{x - 1}}}}{\text{ + C}} \hfill \\ \end{align}

10. $\dfrac{{{\text{2x - 3}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{(2x + 3)}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{2x - 3}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{(2x + 3)}}}}{\text{ = }}\dfrac{{{\text{2x - 3}}}}{{{\text{(x - 1)(x + 1)(2x + 3)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x + 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{(2x + 3)}}}} \hfill \\{\text{2x - 3 = A(x + 1)(2x + 1) + B(x - 1)(2x + 1) + C(x - 1)(x + 1)}} \hfill \\{\text{x = 1 }} \Rightarrow {\text{ A = - 1/10}} \hfill \\{\text{x = - 1 }} \Rightarrow {\text{ - 5 = B( - 2)( - 1) }} \Rightarrow \;{\text{B = 5/2}} \hfill \\ \end{align}

\begin{align}{\text{x = 0 }} \Rightarrow \;{\text{ - 3 = A - B - C }} \Rightarrow \;{\text{C = - 24/5}} \hfill \\\dfrac{{{\text{2x - 3}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{(2x + 3)}}}}{\text{ = }}\dfrac{{{\text{2x - 3}}}}{{{\text{(x - 1)(x + 1)(2x + 3)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{10(x + 1)}}}}{\text{ + }}\dfrac{{\text{5}}}{{{\text{2(x - 1)}}}}{\text{ + }}\dfrac{{{\text{ - 24}}}}{{{\text{5(2x + 3)}}}} \hfill \\\int {\dfrac{{{\text{2x - 3}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{(2x + 3)}}}}} {\text{ = }}\int {\dfrac{{{\text{ - dx}}}}{{{\text{10(x + 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{5dx}}}}{{{\text{2(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{ - 24dx}}}}{{{\text{5(2x + 3)}}}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{10}}}}{\text{log(x + 1) + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log(x - 1) - }}\dfrac{{\text{2}}}{{\text{5}}}{\text{log(2x + 3) + c}} \hfill \\ \end{align}

11. $\mathbf{\dfrac{{{\text{5x}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 4}}} \right){\text{(x - 1)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{5x}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 4}}} \right){\text{(x + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 2)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x + 2)}}}}{\text{ + }}\dfrac{{\text{C}}}{{{\text{(x + 1)}}}} \hfill \\{\text{5x = A(x + 2)(x + 1) + B(x - 2)(x + 1) + c(x + 1)(x + 2)}} \hfill \\{\text{x = 2 }} \Rightarrow \;{\text{10 = A(4)(1) }} \Rightarrow \;{\text{A = 5/2}} \hfill \\{\text{x = - 2 }} \Rightarrow \;{\text{ - 10 = B( - 4)( - 1) }} \Rightarrow {\text{ B = 5/2}} \hfill \\{\text{x = - 1 }} \Rightarrow {\text{ - 5 = C( - 3)(1) }} \Rightarrow {\text{ C = 5/3}} \hfill \\\dfrac{{{\text{5x}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 4}}} \right){\text{(x + 1)}}}}{\text{ = }}\dfrac{{\text{5}}}{{{\text{2(x - 2)}}}}{\text{ - }}\dfrac{{\text{5}}}{{{\text{2(x + 2)}}}}{\text{ + }}\dfrac{{\text{5}}}{{{\text{3(x + 1)}}}} \hfill \\ \end{align}

\begin{align}\dfrac{{{\text{5x}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 4}}} \right){\text{(x + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 2)}}}}{\text{ - }}\dfrac{{\text{B}}}{{{\text{(x + 2)}}}}{\text{ + }}\dfrac{{\text{C}}}{{{\text{(x + 1)}}}} \hfill \\\int {\dfrac{{{\text{5x dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 4}}} \right){\text{(x + 1)}}}}} {\text{ = }}\int {\dfrac{{{\text{5dx}}}}{{{\text{2(x - 2)}}}}} {\text{ - }}\int {\dfrac{{{\text{5dx}}}}{{{\text{2(x + 2)}}}}} {\text{ + }}\int {\dfrac{{{\text{5dx}}}}{{{\text{3(x + 1)}}}}} \hfill \\{\text{ = }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log(x - 2) - }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log(x + 2) + }}\dfrac{{\text{5}}}{{\text{3}}}{\text{log(x + 1) + c}} \hfill \\{\text{ = }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log}}\left( {\dfrac{{{\text{x - 2}}}}{{{\text{x + 2}}}}} \right){\text{ + }}\dfrac{{\text{5}}}{{\text{3}}}{\text{log(x + 1) + c}} \hfill \\ \end{align}

12. $\mathbf{\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + x + 1}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)}}}$

उत्तर: $\dfrac{{{{\text{x}}^{\text{3}}}{\text{ + x + 1}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)}}{\text{ = x + }}\dfrac{{{\text{2x - 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{\text{ = x + }}\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}$

\begin{align}\int {\dfrac{{\left( {{{\text{x}}^{\text{3}}}{\text{ + x + 1}}} \right){\text{dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)}}} {\text{ = }}\int {\text{x}} {\text{dx - }}\int {\dfrac{{{\text{2xdx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}} {\text{ - }}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - log}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{\left| {{\text{x - 1}}} \right|}}{{\left| {{\text{x + 1}}} \right|}}{\text{ + c}} \hfill \\ \end{align}

13. $\mathbf{\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(1 - x)}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(1 - x)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(1 - x)}}}}{\text{ + }}\dfrac{{{\text{Bx + C}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\{\text{2 = A}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + (Bx + C)(1 - x)}} \hfill \\{\text{2 = A}}\left( {{{\text{x}}^{\text{2}}}} \right){\text{ + A + Bx - B}}{{\text{x}}^{\text{2}}}{\text{ + C - Cx}} \hfill \\ \end{align}

\begin{align}{\text{2 = (A - B)}}{{\text{x}}^{\text{2}}}{\text{ + (B - C)x + (A + C)}} \hfill \\{\text{A + C = 2 }} \Rightarrow {\text{ A = 2 - C}} \hfill \\{\text{A - B = 0 }} \Rightarrow \;{\text{A = B}} \hfill \\{\text{B - C = 0 }} \Rightarrow {\text{ B = C}} \hfill \\{\text{C = 1 = A = B}} \hfill \\\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(1 - x)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(1 - x)}}}}{\text{ + }}\dfrac{{{\text{x + 1}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{(1 - x)}}}}{\text{ + }}\dfrac{{\text{x}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + }}\dfrac{{\text{1}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\\int {\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{(1 - x)}}}}} {\text{ = }}\int {\dfrac{{{\text{dx}}}}{{{\text{(1 - x)}}}}} {\text{ + }}\int {\dfrac{{{\text{xdx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{ - }}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \hfill \\{\text{ = log(1 - x) + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x) + c}} \hfill \\ \end{align}

14. $\mathbf{\dfrac{{{\text{3x - 1}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{3x - 1}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x + 2)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}} \hfill \\{\text{3x - 1 = A(x + 2) + B}} \hfill \\{\text{x = - 2 }} \Rightarrow \;{\text{3( - 2) - 1 = B }} \Rightarrow \;{\text{B = - 7}} \hfill \\ \end{align}

\begin{align}{\text{x = 0 }} \Rightarrow {\text{ - 1 = A(2) - 7 }} \Rightarrow {\text{ A = 3}} \hfill \\\dfrac{{{\text{3x - 1}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{3}}}{{{\text{(x + 2)}}}}{\text{ + }}\dfrac{{{\text{ - 7}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}} \hfill \\\int {\dfrac{{{\text{(3x - 1)dx}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}}} {\text{ = }}\int {\dfrac{{{\text{3dx}}}}{{{\text{(x + 2)}}}}} {\text{ - }}\int {\dfrac{{{\text{7dx}}}}{{{{{\text{(x + 2)}}}^{\text{2}}}}}} \hfill \\{\text{ = 3log(x + 2) - }}\dfrac{{\text{7}}}{{{\text{(x + 2)}}}}{\text{ + c}} \hfill \\ \end{align}

15. $\mathbf{\dfrac{{\text{1}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}{\text{ = }}\dfrac{{\text{1}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(x - 1)(x + 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x + 1)}}}}{\text{ + }}\dfrac{{{\text{Cx + D}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\{\text{1 = A(x + 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + B(x - 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + (Cx + D)(x - 1)(x + 1)}} \hfill \\{\text{1 = (A + B + C)}}{{\text{x}}^{\text{3}}}{\text{ + ( - A + B + D)}}{{\text{x}}^{\text{2}}}{\text{ + (A + B - C)x + ( - A + B + D)}} \hfill \\{\text{A + B + C = 0}} \hfill \\{\text{ - A + B + D = 0}} \hfill \\{\text{A + B - C = 0}} \hfill \\{\text{ - A + B + D = 1}} \hfill \\{\text{A = - 1/4, B = 1/4, C = 0, D = - 1/2}} \hfill \\ \end{align}

\begin{align}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}{\text{ = }}\dfrac{{\text{1}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(x - 1)(x + 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{4(x - 1)}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{4(x + 1)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}} \hfill \\ \end{align}

\begin{align}\int {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}} {\text{ = }}\int {\dfrac{{{\text{ - dx}}}}{{{\text{4(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{4(x + 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{dx}}}}{{{\text{2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{4}}}{\text{log(x - 1) + }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log(x + 1) - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x) + C}} \hfill \\\end{align}

16. $\mathbf{\dfrac{{\text{1}}}{{{\text{x}}\left( {{{\text{x}}^{\text{n}}}{\text{ + 1}}} \right)}}}$

उत्तर: $\dfrac{{\text{1}}}{{{\text{x}}\left( {{{\text{x}}^{\text{n}}}{\text{ + 1}}} \right)}}{\text{ = }}\dfrac{{{{\text{x}}^{{\text{n - 1}}}}}}{{{{\text{x}}^{\text{n}}}\left( {{{\text{x}}^{\text{n}}}{\text{ + 1}}} \right)}}$

माना ${{\text{x}}^{\text{n}}}{\text{ = t}}$

\begin{align}\Rightarrow \;{\text{n}}{{\text{x}}^{{\text{n - 1}}}}{\text{dx = dt}} \hfill \\\int {\dfrac{{\left. {{{\text{x}}^{{\text{n( - 1}}}}} \right){\text{dx}}}}{{{{\text{x}}^{\text{n}}}\left( {{{\text{x}}^{\text{n}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} \hfill \\\dfrac{{\text{1}}}{{{\text{t(t + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{t + 1}}}} \hfill \\{\text{1 = A(t + 1) + B(t)}} \hfill \\{\text{t = - 1 }} \Rightarrow {\text{ 1 = B( - 1) }} \Rightarrow {\text{ B = - 1}} \hfill \\{\text{t = 0 }} \Rightarrow {\text{ 1 = A(1) }} \Rightarrow \;{\text{A = 1}} \hfill \\\dfrac{{\text{1}}}{{{\text{t(t + 1)}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{t}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{t + 1}}}} \hfill \\ \end{align}

\begin{align}\int {\dfrac{{\left( {{{\text{x}}^{{\text{n - 1}}}}} \right){\text{dx}}}}{{{{\text{x}}^{\text{n}}}\left( {{{\text{x}}^{\text{n}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} {\text{ - }}\dfrac{{\text{1}}}{{\text{n}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t + 1}}}}} \hfill \\{\text{ = (1/n)log(t) - (1/n)log(1 + t) + c}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{n}}}{\text{log}}\left( {{{\text{x}}^{\text{n}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{n}}}{\text{log}}\left( {{\text{1 + }}{{\text{x}}^{\text{n}}}} \right){\text{ + c}} \hfill \\ \end{align}

17. $\mathbf{\dfrac{{{\text{Cos(x)}}}}{{{\text{(1 - sin(x))(2 - sin(x))}}}}}$

उत्तर: माना ${\text{sin(x) = t }} \Rightarrow {\text{ cos(x)dx = dt}}$

$\int {\dfrac{{{\text{cos(x)dx}}}}{{{\text{(1 - sin(x))(2 - sin(x))}}}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{(1 - t)(2 - t)}}}}}$

दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{1}}}{{{\text{(1 - t)(2 - t)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(1 - t)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(2 - t)}}}} \hfill \\{\text{1 = A(2 - t) + B(1 - t)}} \hfill \\{\text{t = 1 }} \Rightarrow \;{\text{1 = A(1)}} \Rightarrow {\text{ A = 1}} \hfill \\{\text{t = 2 }} \Rightarrow \;{\text{1 = B( - 1) }} \Rightarrow \;{\text{B = - 1}} \hfill \\ \dfrac{{\text{1}}}{{{\text{(1 - t)(2 - t)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(1 - t)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{(2 - t)}}}} \hfill \\\int {\dfrac{{\text{1}}}{{{\text{(1 - t)(2 - t)}}}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{(1 - t)}}}}} {\text{ - }}\int {\dfrac{{{\text{dt}}}}{{{\text{(2 - t)}}}}} \hfill \\{\text{ = - log(1 - t) + log(2 - t) + c}} \hfill \\ \end{align}

\begin{align}{\text{ = - log(1 - t) + log(2 - t) + c}} \hfill \\{\text{ = log}}\left( {\dfrac{{{\text{2 - t}}}}{{{\text{1 - t}}}}} \right){\text{ + c}} \hfill \\{\text{ = log}}\left( {\dfrac{{{\text{2 - sin(x)}}}}{{{\text{1 - sin(x)}}}}} \right){\text{ + c}} \hfill \\ \end{align}

18. $\mathbf{\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 2}}} \right)}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}}$

उत्तर: इसमे ${{\text{x}}^{\text{2}}}$ को ${\text{t}}$ रखने पर

$\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 2}}} \right)}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}{\text{ = }}\dfrac{{{\text{(t + 1)(t + 2)}}}}{{{\text{(t + 3)(t + 4)}}}}{\text{ = }}\dfrac{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 3t + 2}}} \right)}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}$

इस परीमी मे ऊपर और नीचे पावर बराबर है तो भाग लेने पर

\begin{align}\dfrac{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 3t + 2}}} \right)}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = 1 - }}\dfrac{{{\text{(4t + 10)}}}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = 1 - }}\dfrac{{{\text{2(2t + 5)}}}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}} \hfill \\\dfrac{{{\text{(2t + 5)}}}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = }}\dfrac{{{\text{(4t + 10)}}}}{{{\text{(t + 4)(t + 3)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(t + 3)}}}}{\text{ + }}\dfrac{{{\text{(B)}}}}{{{\text{(t + 4)}}}} \hfill \\ {\text{2t + 5 = A(t + 4) + B(t + 3)}} \hfill \\{\text{t = - 3 }} \Rightarrow {\text{ - 6 + 5 = A(1) }} \Rightarrow {\text{ A = - 1}} \hfill \\{\text{t = - 4 }} \Rightarrow {\text{ - 8 + 5 = B( - 1) }} \Rightarrow {\text{ B = 3}} \hfill \\\dfrac{{{\text{(2t + 5)}}}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = }}\dfrac{{{\text{(4t + 10)}}}}{{{\text{(t + 4)(t + 3)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{{\text{(t + 3)}}}}{\text{ + }}\dfrac{{\text{3}}}{{{\text{(t + 4)}}}} \hfill \\\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 2}}} \right)}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}{\text{ = }}\dfrac{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 3t + 2}}} \right)}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = 1 - }}\dfrac{{{\text{(4t + 10)}}}}{{\left( {{{\text{t}}^{\text{2}}}{\text{ + 7t + 12}}} \right)}}{\text{ = 1 - [}}\dfrac{{{\text{ - 2}}}}{{{\text{(t + 3)}}}}{\text{ + }}\dfrac{{\text{6}}}{{{\text{(t + 4)}}}}] \hfill \\ \end{align}

\begin{align}{\text{ = 1 + }}\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}{\text{ - }}\dfrac{{\text{6}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}} \hfill \\\int {\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 2}}} \right){\text{dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}} {\text{ = }}\int {\left[ {{\text{1 + }}\dfrac{{\text{2}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}{\text{ - }}\dfrac{{\text{6}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}} \right]} {\text{ dx}} \hfill \\{\text{ = }}\int {\text{d}} {\text{x + }}\int {\dfrac{{{\text{ - 2dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}} {\text{ - }}\int {\dfrac{{{\text{6dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}} \hfill \\{\text{ = }}\int {\text{d}} {\text{x + 2}}\int {\dfrac{{{\text{dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}} {\text{ - 6}}\int {\dfrac{{{\text{dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)}}} \hfill \\{\text{ = x + }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\sqrt {\text{3}} }}} \right){\text{ - }}\dfrac{{\text{6}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ + C}} \hfill \\\left[ {\int {\dfrac{{{\text{dx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{a}}}} \right){\text{ + C}}} \right] \hfill \\{\text{ = x + }}\dfrac{{\text{2}}}{{\sqrt {\text{3}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\sqrt {\text{3}} }}} \right){\text{ - 3ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ + C}} \hfill \\ \end{align}

19. $\mathbf{\dfrac{{{\text{2x}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}{{\text{x}}^{\text{2}}}{\text{ = t}} \hfill \\{\text{2x dx = dt}} \hfill \\\int {\dfrac{{{\text{2xdx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{(t + 1)(t + 3)}}}}} \hfill \\\dfrac{{\text{1}}}{{{\text{(t + 1)(t + 3)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(t + 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(t + 3)}}}} \hfill \\ \end{align}

\begin{align}{\text{1 = A(t + 3) + B(t + 1)}} \hfill \\{\text{t = - 1 }} \Rightarrow {\text{ 1 = A(2) }} \Rightarrow {\text{ A = 1/2}} \hfill \\{\text{t = - 3 }} \Rightarrow {\text{ 1 = B( - 2) }} \Rightarrow {\text{ B = - 1/2}} \hfill \\ \end{align}

$\dfrac{{\text{1}}}{{{\text{(t + 1)(t + 3)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{2(t + 1)}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{2(t + 3)}}}}$

$\int {\dfrac{{{\text{2xdx}}}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 3}}} \right)}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{(t + 1)(t + 3)}}}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{2(t + 1)}}}}} {\text{ - }}\int {\dfrac{{{\text{dt}}}}{{{\text{2(t + 3)}}}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(t + 1) - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(t + 3)}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {\dfrac{{{\text{t + 1}}}}{{{\text{t + 3}}}}} \right){\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 3}}}}} \right){\text{ + C}} \hfill \\ \end{align}

20. $\mathbf{\dfrac{{\text{1}}}{{{\text{x}}\left( {{{\text{x}}^{\text{4}}}{\text{ - 1}}} \right)}}}$

उत्तर: $\dfrac{{{{\text{x}}^{\text{3}}}}}{{{{\text{x}}^{\text{4}}}\left( {{{\text{x}}^{\text{4}}}{\text{ - 1}}} \right)}}$

\begin{align}{{\text{x}}^{\text{4}}}{\text{ = t}} \hfill \\{\text{4}}{{\text{x}}^{\text{3}}}{\text{ dx = dt}} \hfill \\\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{dx}}}}{{{{\text{x}}^{\text{4}}}\left({{{\text{x}}^{\text{4}}}{\text{ - 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t - 1)}}}}} \hfill \\ \end{align}

\begin{align}\begin{array}{*{20}{c}}{\dfrac{{\text{1}}}{{{\text{t(t - 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(t - 1)}}}}}&{} \\ {{\text{1 = A(t - 1) + B(t)}}}&{} \\ {{\text{t = 0 }} \Rightarrow {\text{ }}}&{{\text{1 = A( - 1) }} \Rightarrow {\text{ A = - 1}}} \\ {{\text{ t = 1 }} \Rightarrow {\text{ 1 = B }} \Rightarrow {\text{ B = 1}}}&{} \\ {\dfrac{{\text{1}}}{{{\text{t(t - 1)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(t - 1)}}}}}&{} \end{array}{\text{1 = A( - 1) }} \Rightarrow {\text{ A = - 1}} \hfill \\\dfrac{{\text{1}}}{{{\text{t(t - 1)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(t - 1)}}}}\int {\dfrac{{{{\text{x}}^{\text{3}}}{\text{dx}}}}{{{{\text{x}}^{\text{4}}}\left({{{\text{x}}^{\text{4}}}{\text{ - 1}}} \right)}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t - 1)}}}}} {\text{ = - }}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} {\text{ + }}\int {\dfrac{{{\text{dt}}}}{{{\text{(t - 1)}}}}} \hfill \\{\text{ = - }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log(t) + }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log(t - 1) + c}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log}}\left( {\dfrac{{{\text{t - 1}}}}{{\text{t}}}} \right){\text{ + c}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}{\text{log}}\left( {\dfrac{{{{\text{x}}^{\text{4}}}{\text{ - 1}}}}{{{{\text{x}}^{\text{4}}}}}} \right){\text{ + c}} \hfill \\ \end{align}

21. $\mathbf{\dfrac{{\text{1}}}{{{{\text{e}}^{\text{x}}}{\text{ - 1}}}}}$

उत्तर: $\dfrac{{\text{1}}}{{{{\text{e}}^{\text{x}}}{\text{ - 1}}}}{\text{ = }}\dfrac{{{{\text{e}}^{\text{x}}}}}{{{{\text{e}}^{\text{x}}}\left( {{{\text{e}}^{\text{x}}}{\text{ - 1}}} \right)}}$

\begin{align}{{\text{e}}^{\text{x}}}{\text{ = t}} \hfill \\{{\text{e}}^{\text{x}}}{\text{dx = dt}} \hfill \\\int {\dfrac{{{\text{dx}}}}{{{{\text{e}}^{\text{x}}}{\text{ - 1}}}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t - 1)}}}}} \hfill \\\dfrac{{\text{1}}}{{{\text{t(t - 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(t - 1)}}}} \hfill \\ \end{align}

\begin{align}{\text{1 = A(t - 1) + B(t)}} \hfill \\{\text{t = 0 }} \Rightarrow {\text{ 1 = A( - 1) }} \Rightarrow \;{\text{A = - 1}} \hfill \\{\text{t = 1 }} \Rightarrow \;{\text{1 = B }} \Rightarrow {\text{ B = 1}} \hfill \\\dfrac{{\text{1}}}{{{\text{t(t - 1)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(t - 1)}}}} \hfill \\\int {\dfrac{{{\text{dx}}}}{{{{\text{e}}^{\text{x}}}{\text{ - 1}}}}} {\text{ = }}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t - 1)}}}}} {\text{ = - }}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} {\text{ + }}\int {\dfrac{{{\text{dt}}}}{{{\text{(t - 1)}}}}} \hfill \\{\text{ = - log(t) + log(t - 1) + c}} \hfill \\{\text{ = log}}\left( {\dfrac{{{\text{t - 1}}}}{{\text{t}}}} \right){\text{ + c}} \hfill \\{\text{ = log}}\left( {\dfrac{{{{\text{e}}^{\text{x}}}{\text{ - 1}}}}{{{{\text{e}}^{\text{x}}}}}} \right){\text{ + c}} \hfill \\ \end{align}

प्रश्न 22 एवं 23 मे सही उत्तर का चयन कीजिए:

22. $\mathbf{\int {\dfrac{{{\text{xdx}}}}{{{\text{(x - 1)(x - 2)}}}}} }$ बराबर है।

1. $\mathbf{{\text{log}}\dfrac{{{\text{|(x - 1)}}{{\text{|}}^{\text{2}}}}}{{{\text{|x - 2|}}}}}$

2. $\mathbf{{\text{log}}\dfrac{{{{{\text{(x - 2)}}}^{\text{2}}}}}{{{\text{(x - 1)}}}}{\text{ + C}}}$

3. $\mathbf{{\text{log}}{\left( {\dfrac{{{\text{|x - 1|}}}}{{{\text{|x - 2|}}}}} \right)^{\text{2}}}{\text{ + C}}}$

4. $\mathbf{{\text{log|(x - 1)(x - 2)| + C}}}$

उत्तर: दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{{\text{xdx}}}}{{{\text{(x - 1)(x - 2)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x - 2)}}}} \hfill \\{\text{x = A(x - 2) + B(x - 1)}} \hfill \\{\text{x = 2 }} \Rightarrow {\text{ 2 = B(1) }} \Rightarrow {\text{ B = 2}} \hfill \\{\text{x = 1 }} \Rightarrow {\text{ - 1 = A(1) }} \Rightarrow {\text{ A = - 1}} \hfill \\ \end{align}

\begin{align}\dfrac{{{\text{xdx}}}}{{{\text{(x - 1)(x - 2)}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{(x - 1)}}}}{\text{ + }}\dfrac{{\text{2}}}{{{\text{(x - 2)}}}} \hfill \\\int {\dfrac{{{\text{xdx}}}}{{{\text{(x - 1)(x - 2)}}}}} \; = \;\int {\dfrac{{{\text{ - dx}}}}{{{\text{(x - 1)}}}}} {\text{ + }}\int {\dfrac{{{\text{2dx}}}}{{{\text{(x - 2)}}}}} \hfill \\{\text{ = - log(x - 1) + 2log(x - 2) + c}} \hfill \\{\text{ = log}}\dfrac{{{{{\text{(x - 2)}}}^{\text{2}}}}}{{{\text{(x - 1)}}}}{\text{ + c}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

23. $\mathbf{\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} }$ बराबर है।

1. $\mathbf{{\text{log|x| - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right|{\text{ + C}}}$

2. $\mathbf{{\text{log|x| + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right|}$

3. $\mathbf{{\text{ - log|x| + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right|{\text{ + C}}}$

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{log|x| + log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right|{\text{ + C}}}$

उत्तर: $\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\int {\dfrac{{{\text{xdx}}}}{{{{\text{x}}^{\text{2}}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{2xdx}}}}{{{{\text{x}}^{\text{2}}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}}$

\begin{align}{{\text{x}}^{\text{2}}}{\text{ = t}} \hfill \\{\text{(2x)dx = dt}} \hfill \\\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{2xdx}}}}{{{{\text{x}}^{\text{2}}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} \hfill \\ \end{align}

दिया हुआ समाकल्य एक उचित परीमी फलन है इसलिए आंशिक भिन्नो का उपयोग करते हुए हम इसे इस तरह अभिव्यक्त करते है:

\begin{align}\dfrac{{\text{1}}}{{{\text{t(t + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{t}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{t + 1}}}} \hfill \\{\text{1 = A(t + 1) + B(t)}} \hfill \\{\text{t = - 1 }} \Rightarrow \;{\text{1 = B( - 1) }} \Rightarrow \;{\text{B = - 1}} \hfill \\ \end{align}

${\text{t = 0 }} \Rightarrow \;{\text{1 = A(1) }} \Rightarrow \;{\text{A = 1}}$

\begin{align}\dfrac{{\text{1}}}{{{\text{t(t + 1)}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{t}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{t + 1}}}} \hfill \\\int {\dfrac{{{\text{dx}}}}{{{\text{x}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{2xdx}}}}{{{{\text{x}}^{\text{2}}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{dt}}}}{{{\text{t(t + 1)}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} {\text{ - }}\dfrac{{{\text{dt}}}}{{{\text{t + 1}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(t) - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log(t + 1) + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{\text{t}}}{{{\text{(t + 1)}}}}{\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{\left( {{{\text{x}}^{\text{2}}}} \right)}}{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{\text{ + c}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + c}} \hfill \\{\text{ = log(x) - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ + c}} \hfill \\ \end{align}

अतः विकल्प (a) सही है।

### प्रश्नावली 7.6

1 से 22 तक प्रश्नों के फलनों का समाकलन कीजिए।

1. $\mathbf{{\text{xsinx}}}$

उत्तर: ${\text{I = }}\int {\text{x}} {\text{ sinx dx}}$

${\text{x}}$ को प्रथम और ${\text{sin x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = x}}\int {{\text{sin}}} {\text{xdx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{x}}} \right)\int {{\text{sin}}} {\text{xdx}}} \right\}} {\text{dx}} \hfill \\{\text{ = x( - cosx) - }}\int {\text{1}} {{ \times ( - cosx)dx}} \hfill \\{\text{ = - x cosx + sinx + C}} \hfill \\ \end{align}

2. $\mathbf{{\text{xsin3x}}}$

उत्तर: ${\text{I = }}\int {\text{x}} {\text{ sin3x dx}}$

${\text{x}}$ को प्रथम और ${\text{sin 3x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = x}}\int {{\text{sin}}} {\text{3xdx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{x}}} \right)\int {{\text{sin}}} {\text{3xdx}}} \right\}} {\text{dx}} \hfill \\{\text{ = x}}\left( {{\text{ - }}\dfrac{{{\text{cos3x}}}}{{\text{3}}}} \right){\text{ - }}\int {\text{1}} {{ \times }}\left( {{\text{ - }}\dfrac{{{\text{cos3x}}}}{{\text{3}}}} \right){\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{\text{ - xcos3x}}}}{{\text{3}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{3}}}\int {{\text{cos}}} {\text{3x dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{\text{ - xcos3x}}}}{{\text{3}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{9}}}{\text{sin3x + C}} \hfill \\ \end{align}

3. $\mathbf{{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}}$

उत्तर: ${\text{I = }}\int {{{\text{x}}^{\text{2}}}} {{\text{e}}^{\text{x}}}{\text{ dx}}$

${{\text{x}}^2}$ को प्रथम और ${{\text{e}}^{\text{x}}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

${\text{I = }}{{\text{x}}^{\text{2}}}\int {{{\text{e}}^{\text{x}}}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^{\text{2}}}} \right)\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right\}} {\text{dx}}$

\begin{align}{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - }}\int {\text{2}} {\text{x}}{{\text{e}}^{\text{x}}}{\text{ dx}} \hfill \\{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - 2}}\int {\text{x}} {{ \times }}{{\text{e}}^{\text{x}}}{\text{ dx}} \hfill \\{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - 2}}\left[ {{\text{x}}{\text{.}}\int {{{\text{e}}^{\text{x}}}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{x}}} \right)\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right\}} {\text{dx}}} \right] \hfill \\{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - 2}}\left[ {{\text{x}}{{\text{e}}^{\text{x}}}{\text{ - }}\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - 2}}\left[ {{\text{x}}{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{\text{x}}}} \right] \hfill \\{\text{ = }}{{\text{x}}^{\text{2}}}{{\text{e}}^{\text{x}}}{\text{ - 2x}}{{\text{e}}^{\text{x}}}{\text{ + 2}}{{\text{e}}^{\text{x}}}{\text{ + C}} \hfill \\{\text{ = }}{{\text{e}}^{\text{x}}}\left( {{{\text{x}}^{\text{2}}}{\text{ - 2x + 2}}} \right){\text{ + C}} \hfill \\ \end{align}

4. $\mathbf{{\text{xlogx}}}$

उत्तर: ${\text{I = }}\int {{\text{x }}} {\text{logx dx}}$

${\text{log x}}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = logx}}\int {\text{x}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)\int {\text{x}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{{ = logx \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{logx}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{\text{x}}}{{\text{2}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{logx}}}}{{\text{2}}}{\text{-}}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}{\text{ + C}} \hfill \\ \end{align}

5. $\mathbf{{\text{xlog2x}}}$

उत्तर: ${\text{I = }}\int {\text{x}} {\text{ log2x dx}}$

${\text{log 2x}}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = log2x}}\int {\text{x}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{log2x}}} \right)\int {\text{x}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{{ = log2x \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{\text{2}}}{{{\text{2x}}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{log2x}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{\text{x}}}{{\text{2}}}} {\text{ dx}} \hfill \\ \end{align}

${\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}{\text{log2x}}}}{{\text{2}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}{\text{ + C}}$

6. $\mathbf{{{\text{x}}^{\text{2}}}{\text{logx}}}$

उत्तर: ${\text{I = }}\int {{{\text{x}}^{\text{2}}}} {\text{ logx dx}}$

${\text{log x}}$ को प्रथम और ${{\text{x}}^2}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

${\text{I = logx}}\int {{{\text{x}}^{\text{2}}}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)\int {{{\text{x}}^{\text{2}}}} {\text{dx}}} \right\}} {\text{dx}}$

\begin{align}{{ = logx \times }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ - }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{ - }}\int {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{3}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{9}}}{\text{ + C}} \hfill \\ \end{align}

7. $\mathbf{{\text{xsi}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$

उत्तर: ${\text{I = }}\int {{\text{x }}} {\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x dx}}$

${\text{si}}{{\text{n}}^{ - 1}}{\text{x}}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\int {\text{x}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {\text{x}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right){\text{ - }}\int {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{ - }}{{\text{x}}^{\text{2}}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left\{ {\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}{\text{ - }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left\{ {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ - }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\left\{ {\int {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } {\text{dx - }}\int {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}}} \right\} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\left\{ {\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - si}}{{\text{n}}^{{\text{ - 1}}}}{\text{xdx}}} \right\}{\text{ + C}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ + }}\dfrac{{\text{x}}}{{\text{4}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{xdx + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\left( {{\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right){\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + }}\dfrac{{\text{x}}}{{\text{4}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + C}} \hfill \\ \end{align}

8. $\mathbf{{\text{xta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$

उत्तर: ${\text{I = }}\int {{\text{x }}} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x dx}}$

${\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\int {\text{x}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {\text{x}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right){\text{ - }}\int {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left\{ {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left\{ {{\text{1 - }}\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{{ \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{\text{x - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ + c}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + C}} \hfill \\{\text{ = }}\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x-}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + C}} \hfill \\ \end{align}

9. $\mathbf{{\text{xco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}$

उत्तर: ${\text{I = }}\int {{\text{x }}} {\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x dx}}$

${\text{co}}{{\text{s}}^{ - 1}}{\text{x}}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}\int {\text{x}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {\text{x}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right){\text{ - }}\int {\dfrac{{{\text{ - 1}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{1 - }}{{\text{x}}^{\text{2}}}{\text{ - 1}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{ dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left\{ {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\left( {\dfrac{{{\text{ - 1}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right)} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} } {\text{dx - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\left( {\dfrac{{{\text{ - 1}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right)} {\text{dx}} \hfill \\ \end{align}

${\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}{\text{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\text{2}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x + C}}$

${\text{ = }}\dfrac{{\left( {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \right)}}{{\text{2}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}\left( {\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right){\text{ + C}}$

10. $\mathbf{{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}}$

उत्तर: ${\text{I = }}\int {{{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)}^{\text{2}}}{\text{dx}}}$

${{\text{(si}}{{\text{n}}^{ - 1}}{\text{x)}}^2}$ को प्रथम और $1$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = }}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}\int {\text{1}} {\text{ dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)}^{\text{2}}}} \right)\int {\text{1}} {\text{ dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = }}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{{ \times x - }}\int {\dfrac{{{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {{ \times x dx}} \hfill \\{\text{ = x}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{\text{ + }}\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{x}}\left( {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} \right){\text{dx}} \hfill \\{\text{ = x}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{\text{ + }}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\int {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}}} \right.\left. {{\text{ - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}}} \right\}} {\text{dx}}} \right] \hfill \\{\text{ = x}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{\text{ + }}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{{x \times 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ - }}\int {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {{ \times 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} \right] \hfill \\{\text{ = x}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{\text{ + 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\int {\text{2}} {\text{dx}} \hfill \\{\text{ = x}}{\left( {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)^{\text{2}}}{\text{ + 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {{ \times si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - 2x + C}} \hfill \\ \end{align}

11. $\mathbf{\dfrac{{{\text{xco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{xco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{ dx}}$

${\text{I = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\int {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{ dx}}$

${\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}$ को प्रथम और $\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\left[ {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{{ \times }}\int {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}}} \right.\left. {{\text{ - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {\dfrac{{{\text{ - 2x}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {\text{dx}}} \right\}} {\text{dx}}} \right] \hfill \\ {\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\left[ {{\text{co}}{{\text{s}}^{{\text{ - 1}}}}{{x \times 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{ - }}\int {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}} {{ \times 2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{dx}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\left[ {{\text{2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x + }}\int {\text{2}} {\text{dx}}} \right] \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\left[ {{\text{2}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x + 2x}}} \right]{\text{ + C}} \hfill \\{\text{ = - }}\left[ {\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {{ \times co}}{{\text{s}}^{{\text{ - 1}}}}{\text{x + x}}} \right]{\text{ + C}} \hfill \\ \end{align}

12. $\mathbf{{\text{xse}}{{\text{c}}^{\text{2}}}{\text{x}}}$

उत्तर: ${\text{I = }}\int {{\text{x }}} {\text{se}}{{\text{c}}^{\text{2}}}{\text{x dx}}$

${\text{x}}$ को प्रथम और ${\text{se}}{{\text{c}}^{\text{2}}}{\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = x}}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{xdx - }}\int {\left\{ {\left\{ {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{x}}} \right\}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {\text{xdx}}} \right\}} {\text{dx}} \hfill \\ {\text{ = x tanx - }}\int {\text{1}} {\text{ \times tanx dx}} \hfill \\{\text{ = x tanx + log|cosx| + C}} \hfill \\ \end{align}

13. $\mathbf{{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}}$

उत्तर: ${\text{I = }}\int {\text{1}} {\text{.ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x dx}}$

${\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}$ को प्रथम और $1$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}\int {\text{1}} {\text{ dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right)\int {{\text{1 }}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{{x \times x - }}\int {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} {{ \times x dx}} \hfill \\ \end{align}

\begin{align}{{ = x \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} {\text{dx}} \hfill \\{{ = x \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right|{\text{ + C}} \hfill \\{{ = x \times ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right){\text{ + C}} \hfill \\ \end{align}

14. $\mathbf{{\text{x(logx}}{{\text{)}}^{\text{2}}}}$

उत्तर: ${\text{I = }}\int {\text{x}} {{\text{(logx)}}^{\text{2}}}{\text{ dx}}$

${{\text{(log x)}}^2}$ को प्रथम और ${\text{x}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

${\text{I = (logx}}{{\text{)}}^{\text{2}}}\int {\text{x}} {\text{dx - }}\int {\left[ {\left\{ {{{\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)}^{\text{2}}}} \right\}\int {\text{x}} {\text{dx}}} \right]} {\text{dx}}$

${\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}{\text{ - }}\left[ {\int {\text{2}} {{logx \times }}\dfrac{{\text{1}}}{{\text{x}}}{{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}}} \right]$

\begin{align}{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}{\text{ - }}\int {\text{x}} {\text{ logx dx}} \hfill \\{\text{I = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}\left[{{\text{logx}}\int {\text{x}} {\text{ dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)\int {\text{x}} {\text{ dx}}} \right\}} {\text{dx}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}{\text{ - }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{logx - }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{dx}}} \right] \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{logx + }}\dfrac{{\text{1}}}{{\text{2}}}\int {{\text{x }}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{{\text{(logx)}}^{\text{2}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{logx + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{4}}}{\text{ + C}} \hfill \\ \end{align}

15. $\mathbf{\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{logx}}}$

उत्तर: ${\text{I = }}\int {\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} {\text{logx dx = }}\int {{{\text{x}}^{\text{2}}}} {\text{logxdx + }}\int {{\text{log}}} {\text{x dx}}$

\begin{align}{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {{{\text{x}}^{\text{2}}}} {\text{ logx dx , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int {{\text{log}}} {\text{x dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {{{\text{x}}^{\text{2}}}} {\text{ logx dx}} \hfill \\ \end{align}

${\text{log x}}$ को प्रथम और ${{\text{x}}^{\text{2}}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = logx}}\int {{{\text{x}}^{\text{2}}}} {\text{dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)\int {{{\text{x}}^{\text{2}}}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{{ = logx \times }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ - }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {{ \times }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{dx}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{ - }}\int {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{3}}}} {\text{ dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{-}}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{9}}}{\text{ + }}{{\text{C}}_{\text{1}}} \hfill \\ {{\text{I}}_{\text{2}}}{\text{ = }}\int {{\text{log}}} {\text{x dx}} \hfill \\ \end{align}

${\text{log x}}$ को प्रथम और $1$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = logx}}\int {\text{1}} {\text{ dx - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{logx}}} \right)\int {\text{1}} {\text{ dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = logx \times x - }}\int {\dfrac{{\text{1}}}{{\text{x}}}} {\text{ \times x dx}} \hfill \\ \end{align}

\begin{align}{\text{ = x logx - }}\int {\text{1}} {\text{ dx}} \hfill \\{\text{ = x logx - x + }}{{\text{C}}_{\text{2}}} \hfill \\{\text{I = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{ - }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{9}}}{\text{ + }}{{\text{C}}_{\text{1}}}{\text{ + x logx - x + }}{{\text{C}}_{\text{2}}} \hfill \\{\text{ = }}\dfrac{{{{\text{x}}^{\text{3}}}{\text{logx}}}}{{\text{3}}}{\text{ -}}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{9}}}{\text{ + x logx - x + }}\left( {{{\text{C}}_{\text{1}}}{\text{ + }}{{\text{C}}_{\text{2}}}} \right) \hfill \\{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{ + x}}} \right){\text{logx - }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{9}}}{\text{ - x + C}} \hfill \\ \end{align}

16. $\mathbf{{{\text{e}}^{\text{x}}}{\text{(sinx + cosx)}}}$

उत्तर: ${\text{I = }}\int {{{\text{e}}^{\text{x}}}} {\text{(sinx + cosx)dx}}$

f(x) = sin x

f’(x) = cos x

$I = \int e^{x} {f(x) + f'(x)}dx\\ \int e^{x} {f(x) + f'(x)}dx = e^x f(x) + c\\I = e^x \sinx +C$

17. $\mathbf{\dfrac{{{\text{x}}{{\text{e}}^{\text{x}}}}}{{{{{\text{(1 + x)}}}^{\text{2}}}}}}$

उत्तर: ${\text{I = }}\int {\dfrac{{{\text{x}}{{\text{e}}^{\text{x}}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} {\text{dx = }}\int {{{\text{e}}^{\text{x}}}} \left\{ {\dfrac{{\text{x}}}{{\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right)}}} \right\}{\text{dx = }}\int {{{\text{e}}^{\text{2}}}} \left\{ {\dfrac{{{\text{1 + x - 1}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right\}{\text{dx}}$

$\int e^{2} [\frac{1}{1+x} - \frac{1}{1+ x^2}]dx\\f(x ) = \frac{1}{1+x}, f'(x) = \frac{1}{(1+x)^2}\\\int \frac{x e^x}{1+x^2}dx = \int e^x ({f(x ) + f'(x)})dx\\=e^x f(x) + C \\\int \frac{x e^x}{1+x^2}dx = \frac{e^x}{1+x^2}+C$

18. $\mathbf{{{\text{e}}^{\text{x}}}\left( {\dfrac{{{\text{1 + sinx}}}}{{{\text{1 + cosx}}}}} \right)}$

उत्तर: ${\text{ = }}{{\text{e}}^{\text{x}}}\left( {\dfrac{{{\text{si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 2sin}}\dfrac{{\text{x}}}{{\text{2}}}{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right)$

\begin{align}{\text{ = }}\dfrac{{{{\text{e}}^{\text{x}}}{{\left( {{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + cos}}\dfrac{{\text{x}}}{{\text{2}}}} \right)}^{\text{2}}}}}{{{\text{2co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{\text{x}}}{\left( {\dfrac{{{\text{sin}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + cos}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{cos}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right)^{\text{2}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{\text{x}}}{\left[{{\text{tan}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 1}}} \right]^{\text{2}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{\text{x}}}\left[ {{\text{1 + ta}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 2tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{\text{x}}}\left[ {{\text{se}}{{\text{c}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ + 2tan}}\dfrac{{\text{x}}}{{\text{2}}}} \right] \hfill \\ \end{align}

$e^x(\frac{1+\sin x}{1 + \cos x})dx = \frac{1}{2}e^x[\sec^2\frac{x}{2} + 2 \tan \frac{x}{2}]\\f(x) = \tan \frac{x}{2}\Rightarrow f'(x) = \frac{1}{2} \sec ^2\frac{x}{2}\\\int e^x {f(x) + f'(x) +C]}\\\int e^x (\frac{1 = \sin x}{1 + \cos x})dx = e^x \tan \frac{x}{2} + C$

19. $\mathbf{{{\text{e}}^{\text{x}}}\left( {\dfrac{{\text{1}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right)}$

उत्तर: ${\text{I = }}\int {{{\text{e}}^{\text{x}}}} \left[ {\dfrac{{\text{1}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right]{\text{dx}}$

$f(x) = \frac{1}{x}\Rightarrow f'(x) = \frac{-1}{x^2}\\\int e^x {f(x) + f'(x)}dx = e^x f(x) + C \\I = \frac{e^x}{x} +C$

20. $\mathbf{\dfrac{{{\text{(x - 3)}}{{\text{e}}^{\text{x}}}}}{{{\text{x - 1}}}}}$

उत्तर: $\int {{{\text{e}}^{\text{x}}}} \left\{ {\dfrac{{{\text{x - 3}}}}{{{{{\text{(x - 1)}}}^{\text{3}}}}}} \right\}{\text{dx = }}\int {{{\text{e}}^{\text{x}}}} \left\{ {\dfrac{{{\text{x - 1 - 2}}}}{{{{{\text{(x - 1)}}}^{\text{3}}}}}} \right\}{\text{dx}}$

$\int e^x (\frac{1}{(x-1)^3} - \frac{2}{(x-1)^3})dx \\f(x) =\frac{1}{(x-1)^3}\Rightarrow f'(x) = \frac{-2}{(x-1)^3}\\\int e^x (f(x) + f'(x))dx = e^x f(x) + C\\\int e^x \frac{x-3}{(x-1)^3}dx = \frac{e^x}{(x-1)^2}+C$

21. $\mathbf{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}$

उत्तर: ${\text{I = }}\int {{{\text{e}}^{{\text{2x}}}}} {\text{ sinx dx}}$

${\text{sin x}}$ को प्रथम और ${{\text{e}}^{{\text{2x}}}}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{I = sinx}}\int {{{\text{e}}^{{\text{2x}}}}} {\text{ - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{sinx}}} \right)\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right\}} {\text{dx}} \hfill \\{{I = sinx \times }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{ - }}\int {{\text{cos}}} {{x \times }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{dx}} \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{-}}\dfrac{{\text{1}}}{{\text{2}}}\int {{{\text{e}}^{{\text{2x}}}}} {\text{cosx dx}} \hfill \\ \end{align}

\begin{align}{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{cosx}}\int {{{\text{e}}^{{\text{2x}}}}} {\text{ - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{cosx}}} \right)\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right\}} {\text{dx}}} \right] \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{-}}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{{cosx \times }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{ - }}\int {{\text{( - sinx)}}} {{ \times }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{dx}}} \right] \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{ -}}\dfrac{{\text{1}}}{{\text{2}}}\left[{\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{cosx}}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}} {\text{dx}}} \right] \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{ - }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{cosx}}}}{{\text{4}}}{\text{-}}\dfrac{{\text{1}}}{{\text{4}}}{\text{I}} \hfill \\{\text{I + }}\dfrac{{\text{1}}}{{\text{4}}}{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{ - }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{cosx}}}}{{\text{4}}} \hfill \\ \end{align}

$\dfrac{{\text{5}}}{{\text{4}}}{\text{I=}}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{ - }}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{cosx}}}}{{\text{4}}}$

\begin{align}{\text{I = }}\dfrac{{\text{4}}}{{\text{5}}}\left[ {\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{sinx}}}}{{\text{2}}}{\text{-}}\dfrac{{{{\text{e}}^{{\text{2x}}}}{\text{cosx}}}}{{\text{4}}}} \right]{\text{ + C}} \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{5}}}{\text{[2sinx - cosx] + C}} \hfill \\ \end{align}

22. $\mathbf{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right)}$

उत्तर: ${\text{x}}\;{\text{ = }}\;{{tan\theta }}\; \Rightarrow \;{\text{dx}}\;{\text{ = }}\;{\text{se}}{{\text{c}}^{\text{2}}}{{\theta }}\;{{d\theta }}$

\begin{align}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{{2tan\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{{\theta }}}}} \right){\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}{{(sin2\theta ) = 2\theta }} \hfill \\ {\text{ = }}\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right){\text{ = }}\int {\text{2}} {{\theta \times se}}{{\text{c}}^{\text{2}}}{{\theta d\theta = 2}}\int {{\theta }} {\text{ se}}{{\text{c}}^{\text{2}}}{{\theta d\theta }} \hfill \\ \end{align}

${{\theta }}$ को प्रथम और ${\text{se}}{{\text{c}}^{\text{2}}}{{\theta }}$ को द्वितीय फलन लेते हुए खण्डश: समाकलन से

\begin{align}{\text{2}}\left[ {{{\theta }}\int {{\text{se}}{{\text{c}}^{\text{2}}}} {{\theta d\theta - }}\int {\left\{ {\left( {\dfrac{{\text{d}}}{{{{d\theta }}}}{{\theta }}} \right)\int {{\text{se}}{{\text{c}}^{\text{2}}}} {{\theta d\theta }}} \right\}} {\text{d\theta }}} \right] \hfill \\{\text{ = 2}}\left[ {{{\theta tan\theta - }}\int {{\text{tan}}} {{\theta d\theta }}} \right] \hfill \\ {{ = 2[\theta tan\theta - log|cos\theta |] + C}} \hfill \\{\text{ = 2}}\left[ {{\text{xta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + log}}\left| {\dfrac{{\text{1}}}{{\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} }}} \right|} \right]{\text{ + C}} \hfill \\{\text{ = 2x ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x + 2}}\left[ {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right)} \right]{\text{ + C}} \hfill \\{\text{ = 2x ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x - log}}\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right){\text{ + C}} \hfill \\ \end{align}

प्रश्न 23 एवं 24 मे सही उत्तर का चयन कीजिए:

23. $\mathbf{\int {{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}\;{\text{dx}}} }$ बराबर है।

1. $\mathbf{\dfrac{{\text{1}}}{{\text{3}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ + C}}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ + C}}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\text{3}}}{{\text{e}}^{\text{2}}}{\text{ + C}}}$

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{ + C}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}\;{\text{dx}}}$

\begin{align}{{\text{x}}^{\text{3}}}{\text{ = t }} \Rightarrow {\text{3}}{{\text{x}}^{\text{2}}}{\text{ = dt}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{3}}}\int {{{\text{e}}^{\text{t}}}} {\text{dt}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\left( {{{\text{e}}^{\text{t}}}} \right){\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}{\text{ + C}} \hfill \\ \end{align}

अतः विकल्प (a) सही है।

24. $\mathbf{\int {{{\text{e}}^{\text{x}}}} {\text{sec(1 + tanx)}}}$ बराबर है।

1. $\mathbf{{{\text{e}}^{\text{x}}}{\text{cosx + C}}}$

2. $\mathbf{{{\text{e}}^{\text{x}}}{\text{secx + C}}}$

3.  $\mathbf{{{\text{e}}^{\text{x}}}{\text{sinx + C}}}$

4. $\mathbf{{{\text{e}}^{\text{x}}}{\text{tanx + C}}}$

उत्तर: ${\text{I = }}\int {{{\text{e}}^{\text{x}}}} {\text{secx(1 + tanx)dx = }}\int {{{\text{e}}^{\text{x}}}} {\text{(secx + secx tanx)dx}}$

$f(x) = \sec x \Rightarrow f'(x) = \sec x \tan x\\\int e^x (f(x) + f'(x))dx = e^x f(x) + C \\I = e^x \sec x +C$

अतः विकल्प (b) सही है।

### प्रश्नावली 7.7

1 से 9 तक के प्रश्नों के फलनों का समाकलन कीजिए:

1. $\mathbf{\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}{\text{dx = }}\int {\sqrt {{{{\text{(2)}}}^{\text{2}}}{\text{ - (x}}{{\text{)}}^{\text{2}}}} } {\text{dx}}} }$

सूत्र ${\text{ = }}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}(\dfrac{{\text{x}}}{{\text{a}}}{\text{ + c}})$

का प्रयोग करने पर

\begin{align}{\text{I = }}\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}}{\text{+ }}\dfrac{{\text{4}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}{\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{4 - }}{{\text{x}}^{\text{2}}}{\text{ + 2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\text{a}}}{\text{ + C}}} \hfill \\ \end{align}

2. $\mathbf{\sqrt {{\text{1 - 4}}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{1 - 4}}{{\text{x}}^{\text{2}}}} } {\text{dx = }}\int {\sqrt {{{{\text{(1)}}}^{\text{2}}}{\text{ - (2x}}{{\text{)}}^{\text{2}}}} } {\text{dx}}$

सूत्र ${\text{ = }}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}(\dfrac{{\text{x}}}{{\text{a}}}{\text{ + c}})$

का प्रयोग करने पर

\begin{align}{\text{ - 2x = 2}} \hfill \\{\text{dx = dt}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\sqrt {{{{\text{(1)}}}^{\text{2}}}{\text{ - (t}}{{\text{)}}^{\text{2}}}} } {\text{dt}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{\text{1}}}{{\text{2}}}\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]{\text{ + C}} \hfill \\{\text{ = }}\dfrac{{\text{t}}}{{\text{4}}}\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{2x + C}} \hfill \\{\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 - 4}}{{\text{x}}^{\text{2}}}{\text{+}}\dfrac{{\text{1}}}{{\text{4}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{2x + C}}} \hfill \\ \end{align}

3. $\mathbf{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 6}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 6}}} } {\text{dx = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4 + 2}}} } {\text{dx}}$

${\text{ = }}\int {\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} \right){\text{ + 2}}} } {\text{dx = }}\int {\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (}}\sqrt {\text{2}} {{\text{)}}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

${\text{I = }}\dfrac{{{{{\text{(x + 2)}}}^{\text{2}}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 6}}} {\text{ + }}\dfrac{{\text{2}}}{{\text{2}}}{\text{log}}\left| {{\text{(x + 2) + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 6}}} } \right|{\text{ + C}}$

${\text{ = }}\dfrac{{{{{\text{(x + 2)}}}^{\text{2}}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 6}}} {\text{ + 2log}}\left| {{\text{(x + 2) + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} } \right|{\text{ + C}}$

4. $\mathbf{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 1}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 1}}} } {\text{dx = }}\int {\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} \right){\text{ - 3}}} } {\text{dx}}$

${\text{ = }}\int {\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ - (}}\sqrt {\text{3}} {{\text{)}}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

${\text{ = }}\dfrac{{{{{\text{(x + 2)}}}^{\text{2}}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 1}}} {\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{log}}\left| {{\text{(x + 2) + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x + 1}}} } \right|{\text{ + C}}$

5. $\mathbf{\sqrt {{\text{1 - 4x - }}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{1 - 4x - }}{{\text{x}}^{\text{2}}}} } {\text{dx = }}\int {\sqrt {{\text{1 - }}\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4 - 4}}} \right)} } {\text{dx = }}\int {\sqrt {\left( {\sqrt {{\text{5}}{{\text{)}}^{\text{2}}}} {\text{ - (x + 2}}{{\text{)}}^{\text{2}}}} \right.} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}(\dfrac{{\text{x}}}{{\text{a}}}{\text{ + C)}}$

का प्रयोग करने पर

${\text{ = }}\dfrac{{{{{\text{(x + 2)}}}^{\text{2}}}}}{{\text{2}}}\sqrt {{\text{1 - 4x - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x + 2}}}}{{\sqrt {\text{5}} }}} \right){\text{ + C}}$

6. $\mathbf{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x - 5}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x - 5}}} } {\text{dx = }}\int {\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ + 4x + 4}}} \right){\text{ - 9}}} } {\text{dx = }}\int {\sqrt {{{{\text{(x + 2)}}}^{\text{2}}}{\text{ + (3}}{{\text{)}}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{dx = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

${\text{ = }}\dfrac{{{\text{(x + 2)}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x - 5}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{log}}\left| {{\text{(x + 2) + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 4x - 5}}} } \right|{\text{ + C}}$

7. $\mathbf{\sqrt {{\text{1 + 3x - }}{{\text{x}}^{\text{2}}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{1 + 3x - }}{{\text{x}}^{\text{2}}}} } {\text{dx = }}\int {\sqrt {{\text{1 - }}\left( {{{\text{x}}^{\text{2}}}{\text{ - 3x + }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}} \right)} } {\text{dx}}$

${\text{ = }}\int {\sqrt {\left( {{\text{1 + }}\dfrac{{\text{9}}}{{\text{4}}}} \right){\text{ - }}\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)} } {\text{dx = }}\int {\sqrt {{{\left( {\dfrac{{\sqrt {{\text{13}}} }}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } {\text{ + }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(}}\dfrac{{\text{x}}}{{\text{a}}}{\text{ + C)}}$

का प्रयोग करने पर

\begin{align}{\text{ = }}\dfrac{{\left( {{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}}{{\text{2}}}\sqrt {{\text{1 + 3x - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{{\text{13}}}}{{{\text{4 \times 2}}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{x - }}\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\sqrt {{\text{13}}} }}{{\text{2}}}}}} \right){\text{ + C}} \hfill \\{\text{ = }}\dfrac{{{\text{(2x - 3)}}}}{{\text{4}}}\sqrt {{\text{1 + 3x - }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{{\text{13}}}}{{\text{8}}}{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{2x - 3}}}}{{\sqrt {{\text{13}}} }}} \right){\text{ + C}} \hfill \\ \end{align}

8. $\mathbf{\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} } {\text{dx = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x + }}\dfrac{{\text{9}}}{{\text{4}}}{\text{ - }}\dfrac{{\text{9}}}{{\text{4}}}} } {\text{dx = }}\int {\sqrt {{{\left( {{\text{x + }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {\dfrac{{\text{3}}}{{\text{2}}}} \right)}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{dx = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

\begin{align}{\text{ = }}\dfrac{{\left( {{\text{x + }}\dfrac{{\text{3}}}{{\text{2}}}} \right)}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{log}}\left| {\left( {{\text{x + }}\dfrac{{\text{3}}}{{\text{2}}}} \right){\text{ + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} } \right|{\text{ + C}} \hfill \\{\text{ = }}\dfrac{{{\text{(2x + 3)}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{8}}}{\text{log}}\left| {\left( {{\text{x + }}\dfrac{{\text{3}}}{{\text{2}}}} \right){\text{ + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 3x}}} } \right|{\text{ + C}} \hfill \\ \end{align}

9. $\mathbf{\sqrt {{\text{1 + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{9}}}} }$

उत्तर: ${\text{I = }}\int {\sqrt {{\text{1 + }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{9}}}{\text{dx}}} } {\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\int {\sqrt {{\text{9 + }}{{\text{x}}^{\text{2}}}{\text{dx}}} } {\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\int {\sqrt {{{{\text{(3)}}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{dx}}} }$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{dx = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\left[ {\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 9}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9}}} } \right|} \right]{\text{ + C}} \hfill \\ {\text{ = }}\dfrac{{\text{x}}}{{\text{6}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ + 9}}} {\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 9}}} } \right|{\text{ + C}} \hfill \\ \end{align}

प्रश्न 10 एवं 11 मे सही उत्तर का चयन कीजिए:

10. $\int {\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} } {\text{dx}}$ बराबर है।

1. $\mathbf{\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {\left( {{\text{x + }}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} } \right)} \right|{\text{ + C}}}$

2. $\mathbf{\dfrac{{\text{2}}}{{\text{3}}}{\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right)^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + C}}}$

3. $\mathbf{\dfrac{{\text{2}}}{{\text{3}}}{\text{x}}{\left( {{\text{1 + }}{{\text{x}}^{\text{2}}}} \right)^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + C}}}$

4. $\mathbf{\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{x}}^{\text{2}}}{\text{log}}\left| {\left( {{\text{x + }}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} } \right)} \right|{\text{ + C}}}$

उत्तर: सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{dx = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर

${\text{I = }}\dfrac{{\text{x}}}{{\text{2}}}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} {\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left| {\left( {{\text{x + }}\sqrt {{\text{1 + }}{{\text{x}}^{\text{2}}}} } \right)} \right|{\text{ + C}}$

अतः विकल्प (a) सही है।

11. $\mathbf{\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } {\text{dx}}}$ बराबर है।

1. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{(x - 4)}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} {\text{ + 9log}}\left| {{\text{x - 4 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } \right|{\text{ + C}}}$

2. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{(x + 4)}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} {\text{ + 9log}}\left| {{\text{x + 4 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } \right|{\text{ + C}}}$

3. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{(x - 4)}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} {\text{ - 3}}\sqrt {\text{2}} {\text{log}}\left| {{\text{x - 4 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } \right|{\text{ + C}}}$

4. $\mathbf{\dfrac{{\text{1}}}{{\text{2}}}{\text{(x - 4)}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{log}}\left| {{\text{x - 4 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } \right|{\text{ + C}}}$

उत्तर:  ${\text{I = }}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } {\text{dx = }}\int {\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ - 8x + 16}}} \right){\text{ - 9}}} } {\text{dx = }}\int {\sqrt {{{{\text{(x - 4)}}}^{\text{2}}}{\text{ - (3}}{{\text{)}}^{\text{2}}}} } {\text{dx}}$

सूत्र $\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{dx = }}\dfrac{{\text{x}}}{{\text{2}}}\int {\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } {\text{ - }}\dfrac{{{{\text{a}}^{\text{2}}}}}{{\text{2}}}{\text{log}}\left| {{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} } \right|{\text{ + C}}$

का प्रयोग करने पर ${\text{ = }}\dfrac{{{\text{(x - 4)}}}}{{\text{2}}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} {\text{ - }}\dfrac{{\text{9}}}{{\text{2}}}{\text{log}}\left| {{\text{x - 4 + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 8x + 7}}} } \right|{\text{ + C}}$

अतः विकल्प (d) सही है।

### प्रश्नावली 7.8

योगों की सीमा के रूप मे निम्नलिखित निश्चित समाकलनों का मान ज्ञात कीजिए:

1. $\mathbf{\int_{\text{a}}^{\text{b}} {{\text{x }}} {\text{dx}}}$

उत्तर: $\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{ - a = a, b = b, h = }}\dfrac{{{\text{b - a}}}}{{\text{n}}}{\text{, f(x) = x}}$

\begin{align}{\text{I = }}\int_{\text{a}}^{\text{b}} {\text{x}} {\text{ dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{{[a + (a + h) + (a + 2h) \ldots + (a + (n - 1)h)]}} \hfill \\{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{{[(a + a + a + a \ldots a) + (h + 2h + 3h \ldots (n - 1)h)]}} \hfill \\{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[na + h(1 + 2 + 3 + }}...{\text{ + n(n - 1)]}} \hfill \\{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{na + h}}\left\{ {\dfrac{{{\text{(n - 1)(n)}}}}{{\text{2}}}} \right\}} \right]{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{n}}}{{\text{n}}}{\text{[a + }}\left. {\dfrac{{{\text{(n - 1)h}}}}{{\text{2}}}} \right] \hfill \\{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \left[ {{\text{a + }}\dfrac{{{\text{(n - 1)(b - a)}}}}{{{\text{2n}}}}} \right]{\text{ = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \left[ {{\text{a + }}\dfrac{{\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{(b - a)}}}}{{\text{2}}}} \right] \hfill \\ \end{align}

${\text{ = (b - a)}}\left[ {{\text{a + }}\dfrac{{{\text{(b - a)}}}}{{\text{2}}}} \right]{\text{ = (b - a)}}\left[ {\dfrac{{{\text{2a + b - a}}}}{{\text{2}}}} \right]{\text{ = }}\dfrac{{{\text{(b - a)(b + a)}}}}{{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{a}}^{\text{2}}}} \right)$

2. $\mathbf{\int_{\text{a}}^{\text{b}} {{\text{(x + 1)}}} {\text{dx}}}$

उत्तर: $\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{ - a = 0, b = 5, h = }}\dfrac{{{\text{5 - 0}}}}{{\text{n}}}\; = \;\dfrac{5}{{\text{n}}}{\text{, f(x) = x + 1}}$

\begin{align}{\text{I = }}\int_{\text{a}}^{\text{b}} {{\text{(x + 1)}}} {\text{dx = (5 - 0)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{f(0) + f}}\left( {\dfrac{{\text{5}}}{{\text{n}}}} \right){\text{ + }}...{\text{ + f}}\left( {{\text{(n - 1)}}\dfrac{{\text{5}}}{{\text{n}}}} \right)} \right] \hfill \\{\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{1 + }}\left( {\dfrac{{\text{5}}}{{\text{n}}}{\text{ + 1}}} \right){\text{ + }}...{\text{ + f}}\left( {{\text{1 + }}\dfrac{{{\text{5(n - 1)}}}}{{\text{n}}}} \right)} \right] \hfill \\ \end{align}

${\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{(1 + 1 + 1 \ldots 1) + }}\left[ {\dfrac{{\text{5}}}{{\text{n}}}{{ + 2 \times }}\dfrac{{\text{5}}}{{\text{n}}}{\text{ + 3}}\dfrac{{\text{5}}}{{\text{n}}}{\text{ + }}...{\text{(n - 1)}}\dfrac{{\text{5}}}{{\text{n}}}} \right]} \right]$

\begin{align}{\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}\dfrac{{\text{5}}}{{\text{n}}}{\text{\{ 1 + 2 + 3 \ldots (n - 1)\} }}} \right]{\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}\dfrac{{\text{5}}}{{\text{n}}}{{ \times }}\dfrac{{{\text{(n - 1)n}}}}{{\text{2}}}} \right] \hfill \\{\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}\dfrac{{{\text{5(n - 1)}}}}{{\text{2}}}} \right]{\text{ = 5}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{n}}}{{\text{n}}}\left[ {{\text{1 + }}\dfrac{{\text{5}}}{{\text{2}}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right)} \right]{\text{ = 5}}\left[ {{\text{1 + }}\dfrac{{\text{5}}}{{\text{2}}}} \right] \hfill \\{\text{ = 5}}\left[ {\dfrac{{\text{7}}}{{\text{2}}}} \right]{\text{ = }}\dfrac{{{\text{35}}}}{{\text{2}}} \hfill \\ \end{align}

3. $\mathbf{\int_{\text{2}}^{\text{3}} {{{\text{x}}^{\text{2}}}} {\text{dx}}}$

उत्तर: $\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{ - a = 2, b = 3, h = }}\dfrac{{{\text{3 - 2}}}}{{\text{n}}}\;{\text{ = }}\;\dfrac{1}{{\text{n}}}{\text{, f(x) = }}{{\text{x}}^2}$

\begin{align}{\text{I = }}\int_{\text{2}}^{\text{3}} {{{\text{x}}^{\text{2}}}} {\text{dx = (3 - 2)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{f(2) + f}}\left( {{\text{2 + }}\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{ + }}...{\text{ + f(2 + }}} \right.\left. {\left. {{\text{(n - 1)}}\dfrac{{\text{1}}}{{\text{n}}}} \right)} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{{\text{(2)}}}^{\text{2}}}{\text{ + }}{{\left( {{\text{2 + }}\dfrac{{\text{2}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ \ldots + }}{{\left( {\dfrac{{{\text{(n - 1)}}}}{{\text{n}}}} \right)}^{\text{2}}}} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{{\text{(2)}}}^{\text{2}}}{\text{ + }}\left\{ {{{\text{2}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\text{1}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + 2}}{{.2 \times }}\dfrac{{\text{1}}}{{\text{n}}}} \right\}{\text{ + }}...{\text{ + }}\left\{ {{{\text{2}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{{\text{n - 1}}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + }}} \right.} \right.\left. {\left. {{\text{2}}{\text{.2}}\dfrac{{{\text{(n - 1)}}}}{{\text{n}}}} \right\}} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\left( {{{\text{2}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ + }}...{\text{ + }}{{\text{2}}^{\text{2}}}} \right){\text{ + }}\left\{ {{{\left( {\dfrac{{\text{1}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\text{2}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + }}...{\text{ + }}} \right.} \right.\left. {\left. {{{\left( {\dfrac{{{\text{n - 1}}}}{{\text{2}}}} \right)}^{\text{2}}}} \right\}{\text{ + 2}}{\text{.2}}\left\{ {\dfrac{{\text{1}}}{{\text{n}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{n}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{n}}}{\text{ + }}...{\text{ + }}\dfrac{{{\text{(n - 1)}}}}{{\text{n}}}} \right\}} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{4n + }}\dfrac{{\text{1}}}{{{{\text{n}}^{\text{2}}}}}\left\{ {{{\text{1}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ + }}{{\text{3}}^{\text{2}}}{\text{ + }}...{\text{ + (n - 1}}{{\text{)}}^{\text{2}}}} \right\}{\text{ + }}\dfrac{{\text{4}}}{{\text{n}}}{\text{\{ 1 + 2 + }}} \right....{\text{ + (n - 1)\} ]}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{4n + }}\dfrac{{{\text{n}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right)\left( {{\text{2 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right)}}{{\text{6}}}{\text{ + }}\dfrac{{{\text{4n - 4}}}}{{\text{2}}}} \right]{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{4 + }}\dfrac{{\text{1}}}{{\text{6}}}{\text{(1 - }}} \right.\left. {\left. {\dfrac{{\text{1}}}{{\text{n}}}} \right)\left( {{\text{2 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right)\left( {{\text{3 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{ + 2 - }}\dfrac{{\text{2}}}{{\text{n}}}} \right] \hfill \\{\text{ = 4 + }}\dfrac{{\text{2}}}{{\text{6}}}{\text{ + 2 = }}\dfrac{{{\text{19}}}}{{\text{3}}} \hfill \\ \end{align}

4. $\mathbf{\int_{\text{1}}^{\text{4}} {\left( {{{\text{x}}^{\text{2}}}{\text{ - x}}} \right)} {\text{dx}}}$

उत्तर: $\int\limits_{\text{1}}^{\text{4}} {{\text{(}}{{\text{x}}^{\text{2}}}{\text{ - x)dx}}\;{\text{ = }}\;\int\limits_{\text{1}}^{\text{4}} {{{\text{x}}^{\text{2}}}{\text{dx}}\;{\text{ - }}\;\int\limits_{\text{1}}^{\text{4}} {{\text{x}}\;} {\text{dx}}} }$

$\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{a = 1, b = 4, h = }}\dfrac{{{\text{4 - 1}}}}{{\text{n}}}\;{\text{ = }}\;\dfrac{3}{{\text{n}}}{\text{, f(x) = }}{{\text{x}}^2}$

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {{{\text{x}}^{\text{2}}}} {\text{dx , }}{{\text{I}}_{\text{2}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{x}} {\text{dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {{{\text{x}}^{\text{2}}}} {\text{dx = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{{\text{(1)}}}^{\text{2}}}{\text{ + }}{{\left( {{\text{1 + }}\dfrac{{\text{3}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{{1 + 2 \times }}\dfrac{{\text{3}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ \ldots + }}{{\left( {{\text{1 + }}\dfrac{{{\text{(n - 1)3}}}}{{\text{n}}}} \right)}^{\text{2}}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{1}}^{\text{2}}}{\text{ + }}\left\{ {{{\text{1}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\text{3}}}{{\text{n}}}} \right)}^{\text{2}}}{{ + 2 \times }}\dfrac{{\text{3}}}{{\text{n}}}} \right\}{\text{ + }}...{\text{ + }}\left\{ {{{\text{1}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{{\text{(n - 1)3}}}}{{\text{n}}}} \right)}^{\text{2}}}{\text{ + }}} \right.} \right.\left. {\left. {\dfrac{{{{2 \times (n - 1)3}}}}{{\text{n}}}} \right\}} \right] \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\left( {{{\text{1}}^{\text{2}}}{\text{ + }}...{\text{ + }}{{\text{1}}^{\text{2}}}} \right){\text{ + }}{{\left( {\dfrac{{\text{3}}}{{\text{n}}}} \right)}^{\text{2}}}\left\{ {{{\text{1}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ \ldots + (n - 1}}{{\text{)}}^{\text{2}}}} \right\}{\text{ + }}} \right.\left. {{{2 \times }}\dfrac{{\text{3}}}{{\text{n}}}{\text{\{ 1 + 2 + 3 + }}...{\text{(n - 1)\} }}} \right] \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}{{\left( {\dfrac{{\text{3}}}{{\text{n}}}} \right)}^{\text{2}}}\left\{ {\dfrac{{{\text{(n - 1)(n)(2n - 1)}}}}{{\text{6}}}} \right\}{\text{ + }}\dfrac{{\text{6}}}{{\text{n}}}\left\{ {\dfrac{{{\text{(n - 1)(n)}}}}{{\text{2}}}} \right\}} \right] \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}\dfrac{{{\text{9n}}}}{{\text{6}}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{ + }}\dfrac{{{\text{6n - 6}}}}{{\text{2}}}} \right] \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{1 + }}\dfrac{{\text{9}}}{{\text{6}}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{(2 - }}} \right.\left. {\left. {\dfrac{{\text{1}}}{{\text{n}}}} \right){\text{ + 3 - }}\dfrac{{\text{3}}}{{\text{n}}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = 3[1 + 3 + 3] = 21 }} \Rightarrow {\text{ }}{{\text{I}}_{\text{1}}}{\text{ = 21}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{x}} {\text{dx}} \hfill \\{\text{a = 1, b = 4, f(x) = }}{{\text{x}}^{\text{2}}}{\text{, h = }}\dfrac{{{\text{4 - 1}}}}{{\text{n}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{n}}} \hfill \\{\text{ = (4 - 1)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}} \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[1 + (1 + h) + }}...{\text{ + (1 + (n - 1)h)]}} \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{1 + }}\left( {{\text{1 + }}\dfrac{{\text{3}}}{{\text{n}}}} \right){\text{ + }}...{\text{ + }}\left\{ {{\text{1 + (n - 1)}}\dfrac{{\text{3}}}{{\text{n}}}} \right\}} \right] \hfill \\{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[(1 + 1 + }}...{\text{ + 1) + }}\dfrac{{\text{3}}}{{\text{n}}}{\text{(1 + 2 + }}...{\text{(n - 1)}}] \hfill \\{\text{ = }}\;{\text{3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{n + }}\dfrac{{\text{3}}}{{\text{n}}}\left\{ {\dfrac{{{\text{(n - 1)n}}}}{{\text{2}}}} \right\}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = 3}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{1 + }}\dfrac{{\text{3}}}{{\text{2}}}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{n}}}} \right)} \right]{\text{ = 3}}\left[ {{\text{1 + }}\dfrac{{\text{3}}}{{\text{2}}}} \right] \hfill \\{{ = 3 \times }}\dfrac{{\text{5}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ }} \Rightarrow {\text{ }}{{\text{I}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{15}}}}{{\text{2}}} \hfill \\ \end{align}

\begin{align}{\text{I = }}{{\text{I}}_{\text{1}}}{\text{ - }}{{\text{I}}_{\text{2}}} \hfill \\{\text{ = 21 - }}\dfrac{{{\text{15}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{27}}}}{{\text{2}}} \hfill \\ \end{align}

5. $\mathbf{\int_{{\text{ - 1}}}^{\text{1}} {{{\text{e}}^{\text{x}}}} {\text{dx}}}$

उत्तर: $\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{ - a = - 1, b = 1, h = }}\dfrac{{{\text{1 - ( - 1)}}}}{{\text{n}}}\;{\text{ = }}\;\dfrac{{\text{2}}}{{\text{n}}}{\text{, f(x) = }}{{\text{e}}^{\text{x}}}$

\begin{align}{\text{ = (1 + 1)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{f( - 1) + f}}\left( {{\text{ - 1 + }}\dfrac{{\text{2}}}{{\text{n}}}} \right){\text{ + f}}\left( {{{ - 1 + 2 \times }}\dfrac{{\text{2}}}{{\text{n}}}} \right){\text{ \ldots + }}} \right.\left. {{\text{f}}\left( {{\text{ - 1 + }}\dfrac{{{\text{(n - 1)2}}}}{{\text{n}}}} \right)} \right] \hfill \\{\text{ = (2)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{e}}^{{\text{ - 1}}}}{\text{ + }}{{\text{e}}^{\left( {{\text{ - 1 + }}\dfrac{{\text{2}}}{{\text{n}}}} \right)}}{\text{ + }}{{\text{e}}^{\left( {{{ - 1 + 2 \times }}\dfrac{{\text{2}}}{{\text{n}}}} \right)}}{\text{ + }}...{{\text{e}}^{\left( {{\text{ - 1 + (n - 1)}}\dfrac{{\text{2}}}{{\text{n}}}} \right)}}} \right] \hfill \\{\text{ = 2}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{e}}^{{\text{ - 1}}}}\left\{ {{\text{1 + }}\dfrac{{\text{2}}}{{{\text{en}}}}{\text{ + }}\dfrac{{\text{4}}}{{{\text{en}}}}{\text{ + }}\dfrac{{\text{6}}}{{{\text{en}}}}{\text{ + }}...{\text{ + }}{{\text{e}}^{{\text{(n - 1)}}\dfrac{{\text{2}}}{{\text{n}}}}}} \right\}} \right] \hfill \\ \end{align}

6. $\mathbf{\int_{\text{0}}^{\text{4}} {\left( {{\text{x + }}{{\text{e}}^{{\text{2x}}}}} \right)} {\text{dx}}}$

उत्तर: $\int_{\text{a}}^{\text{b}} {{\text{(x)}}} {\text{dx = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}}$

के प्रयोग करने पर

मान लीजिए: ${\text{ - a = - 0, b = 4, h = }}\dfrac{{{\text{4 - 0}}}}{{\text{n}}}\;{\text{ = }}\;\dfrac{{\text{4}}}{{\text{n}}}{\text{, f(x) = x + }}{{\text{e}}^{{\text{2x}}}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{4}} {\left( {{\text{x + }}{{\text{e}}^{{\text{2x}}}}} \right)} {\text{dx = (4 - 0)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + }}...{\text{ + f(a + (n - 1)h)]}} \hfill \\{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\left( {{\text{0 + }}{{\text{e}}^{\text{0}}}} \right){\text{ + }}\left( {{\text{h + }}{{\text{e}}^{{\text{2h}}}}} \right){\text{ + }}\left( {{\text{h + }}{{\text{e}}^{{\text{2}}{\text{.2h}}}}} \right){\text{ + }}...{\text{ + \{ (n - }}} \right.{\text{1)}}\left. {\left. {{\text{h + }}{{\text{e}}^{{\text{2(n - 1)h}}}}} \right\}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{1 + }}\left( {{\text{h + }}{{\text{e}}^{{\text{2h}}}}} \right){\text{ + }}\left( {{\text{h + }}{{\text{e}}^{{\text{2}}{\text{.2h}}}}} \right){\text{ + }}...{\text{ + \{ (n - 1)h + }}} \right.\left. {\left. {{{\text{e}}^{{\text{2(n - 1)h}}}}} \right\}} \right] \hfill \\{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{\{ h + 2h + 3h + }}...{\text{ + (n - 1)h\} + }}\left( {{\text{1 + }}{{\text{e}}^{{\text{2h}}}}{\text{ + }}{{\text{e}}^{{\text{2}}{\text{.2h}}}}} \right)} \right] \hfill \\ \end{align}

\begin{align}{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{\text{h\{ 1 + 2 + \ldots (n - 1)\} + }}\left( {\dfrac{{{{\text{e}}^{{\text{2}}{\text{.2h}}}}{\text{-1}}}}{{{{\text{e}}^{{\text{2}}{\text{.2h}}}}{\text{ - 1}}}}} \right)} \right] \hfill \\{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\dfrac{{{\text{h(n - 1)n}}}}{{\text{2}}}{\text{ + }}\left( {\dfrac{{{{\text{e}}^{{\text{2}}{\text{.2h}}}}{\text{ - 1}}}}{{{{\text{e}}^{{\text{2}}{\text{.2h}}}}{\text{ - 1}}}}} \right)} \right] \hfill \\{\text{ = 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\dfrac{{\text{4}}}{{\text{n}}}{{ \times }}\dfrac{{{\text{(n - 1)n}}}}{{\text{2}}}{\text{ + }}\left({\dfrac{{{{\text{e}}^{\text{h}}}{\text{-1}}}}{{{{\text{e}}^{\dfrac{{\text{8}}}{{\text{n}}}}}{\text{ - 1}}}}} \right)} \right] \hfill \\{\text{ = 4(2) + 4}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{{{\text{e}}^{\text{8}}}{\text{ - 1}}}}{{{\text{(}}\dfrac{{{{\text{e}}^{\dfrac{{\text{8}}}{{\text{n}}}{\text{-1}}}}}}{{\dfrac{{\text{8}}}{{\text{n}}}}}{\text{)8}}}} \hfill \\{\text{ = 8 + }}\dfrac{{{\text{4}}\left( {{{\text{e}}^{\text{8}}}{\text{ - 1}}} \right)}}{{\text{8}}}{\text{ = 8 + }}\dfrac{{\left( {{{\text{e}}^{\text{8}}}{\text{ - 1}}} \right)}}{{\text{2}}} \hfill \\{\text{ = }}\dfrac{{{\text{15 + }}{{\text{e}}^{\text{8}}}}}{{\text{2}}} \hfill \\ \end{align}

### प्रश्नावली 7.9

1 से 20 तक के प्रश्न मे निश्चित समाकलनों का मान ज्ञात कीजिए:

1. $\mathbf{\int_{{\text{ - 1}}}^{\text{1}} {{\text{(x + 1)}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{{\text{ - 1}}}^{\text{1}} {{\text{(x + 1)}}} {\text{dx}}$

\begin{align}{\text{ = }}\left( {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + x}}} \right)_{{\text{ - 1}}}^{\text{1}} \hfill \\{\text{ = }}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 1}}} \right){\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ - 1}}} \right) \hfill \\{\text{ = 2}} \hfill \\ \end{align}

2. $\int_{\text{2}}^{\text{3}} {\dfrac{{\text{1}}}{{\text{x}}}} {\text{dx}}$

उत्तर: ${\text{I = }}\int_{\text{2}}^{\text{3}} {\dfrac{{\text{1}}}{{\text{x}}}} {\text{ dx}}$

\begin{align}{\text{ = [logx]}}_{\text{2}}^{\text{3}} \hfill \\{\text{ = log3 - log2}} \hfill \\{\text{ = log}}\dfrac{{\text{3}}}{{\text{2}}} \hfill \\ \end{align}

3. $\mathbf{\int_{\text{1}}^{\text{2}} {\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - 5}}{{\text{x}}^{\text{2}}}{\text{ + 6x + 9}}} \right)} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{1}}^{\text{2}} {\left( {{\text{4}}{{\text{x}}^{\text{3}}}{\text{ - 5}}{{\text{x}}^{\text{2}}}{\text{ + 6x + 9}}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\left[ {{{4 \times }}\dfrac{{{{\text{x}}^{\text{4}}}}}{{\text{4}}}{{ - 5 \times }}\dfrac{{{{\text{x}}^{\text{3}}}}}{{\text{3}}}{{ + 6 \times }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + 9x}}} \right]_{\text{1}}^{\text{2}} \hfill \\{\text{ = }}\left[ {{{\text{x}}^{\text{4}}}{\text{ - }}\dfrac{{{\text{5}}{{\text{x}}^{\text{3}}}}}{{\text{3}}}{\text{+3}}{{\text{x}}^{\text{2}}}{\text{ + 9x}}} \right]_{\text{1}}^{\text{2}} \hfill \\{\text{ = }}\left( {{{\text{2}}^{\text{4}}}{\text{ - }}{{\text{1}}^{\text{4}}}} \right){\text{ - }}\dfrac{{\text{5}}}{{\text{3}}}\left( {{{\text{2}}^{\text{3}}}{\text{ - }}{{\text{1}}^{\text{3}}}} \right){\text{ + 3}}\left( {{{\text{2}}^{\text{2}}}{\text{ - 1}}} \right){\text{ + 9(2 - 1)}} \hfill \\{\text{ = (16 - 1) - }}\dfrac{{\text{5}}}{{\text{3}}}{\text{(8 - 1) + 3(4 - 1) + 9(1)}} \hfill \\{\text{ = }}\dfrac{{{\text{64}}}}{{\text{3}}} \hfill \\ \end{align}

4. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{sin}}} {\text{2x dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{sin}}} {\text{2x dx}}$

\begin{align}{\text{=}}\dfrac{{\text{1}}}{{\text{2}}}{\text{[cos2x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}\left( {{\text{cos}}\dfrac{{{{2\pi }}}}{{\text{4}}}{\text{ - cos0}}} \right) \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{2}}}{\text{(0 - 1)}} \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}} \hfill \\ \end{align}

5. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{cos}}} {\text{2x dx}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{cos}}} {\text{2x dx}}$

${\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[sin2x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{sin}}\dfrac{{{{2\pi }}}}{{\text{2}}}{\text{ - sin0}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(0 - 0)}} \hfill \\{\text{ = 0}} \hfill \\ \end{align}

6. $\mathbf{\int_{\text{4}}^{\text{5}} {{{\text{e}}^{\text{x}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{4}}^{\text{5}} {{{\text{e}}^{\text{x}}}} {\text{dx = }}\left[ {{{\text{e}}^{\text{x}}}} \right]_{\text{4}}^{\text{5}}$

\begin{align}{\text{ = }}{{\text{e}}^{\text{5}}}{\text{ - }}{{\text{e}}^{\text{4}}} \hfill \\{\text{ = }}{{\text{e}}^{\text{4}}}{\text{(e - 1)}} \hfill \\ \end{align}

7. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{tan}}} {\text{x dx}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{tan}}} {\text{x dx}}$

\begin{align}{\text{ = - [logcosx]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\{\text{ = - }}\left[ {{\text{log cos}}\dfrac{{{\pi }}}{{\text{4}}}{\text{ - log cos0}}} \right] \hfill \\ \end{align}

${\text{ = - }}\left[ {{\text{log}}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{ - log1}}} \right]$${\text{ = log}}\sqrt {\text{2}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log2}}$

8. $\mathbf{\int_{\dfrac{{{\pi }}}{{\text{6}}}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{cosec}}} {\text{x dx}}}$

उत्तर: ${\text{I = }}\int_{\dfrac{{{\pi }}}{{\text{6}}}}^{\dfrac{{\text{\pi }}}{{\text{4}}}} {{\text{cosec}}} {\text{x dx}}$

\begin{align}{\text{ = [log(cosecx - cotx)]}}\dfrac{{{\pi }}}{{\dfrac{{{\pi }}}{{\text{6}}}}} \hfill \\{\text{ = log}}\left[ {{\text{cosec}}\dfrac{{{\pi }}}{{\text{4}}}{\text{ - cot}}\dfrac{{{\pi }}}{{\text{4}}}} \right]{\text{ - log}}\left[ {{\text{cosec}}\dfrac{{\text{\pi }}}{{\text{6}}}{\text{ - cot}}\dfrac{{{\pi }}}{{\text{2}}}} \right] \hfill \\{\text{ = log(}}\sqrt {\text{2}} {\text{ - 1) - log(2 - }}\sqrt {\text{3}} {\text{)}} \hfill \\{\text{ = log}}\dfrac{{\sqrt {\text{2}} {\text{ - 1}}}}{{{\text{2 - }}\sqrt {\text{3}} }} \hfill \\ \end{align}

9.$\mathbf{\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} }}}$

\begin{align}{\text{ = }}\left[ {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 0}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}} \hfill \\ \end{align}

10. $\mathbf{\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}}$

\begin{align}{\text{ = }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{0}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{{\pi }}}{{\text{4}}}{\text{ - 0}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

11. $\mathbf{\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}} }$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ - 1}}}}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{log}}\dfrac{{{\text{x - 1}}}}{{{\text{x + 1}}}}} \right]_{\text{2}}^{\text{3}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{log}}\dfrac{{{\text{3 - 1}}}}{{{\text{3 + 1}}}}{\text{ - log}}\dfrac{{{\text{2 - 1}}}}{{{\text{2 + 1}}}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{log}}\dfrac{{\text{2}}}{{\text{4}}}{\text{ - log}}\dfrac{{\text{1}}}{{\text{3}}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ \times }}\dfrac{{\text{3}}}{{\text{1}}}}\right){\text{=}}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{\text{3}}}{{\text{2}}} \hfill \\ \end{align}

12. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{x dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{x dx}}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{1 + cos2x}}}}{{\text{2}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{dx + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cos2x}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{x + }}\dfrac{{{\text{sin2x}}}}{{\text{2}}}} \right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ + sin}}\dfrac{{{\pi }}}{{\text{2}}}} \right){\text{ - (0 + 0)}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{{\pi }}}{{\text{2}}}{\text{ + 0}}} \right] \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

13. $\mathbf{\int_{\text{2}}^{\text{3}} {\dfrac{{{\text{x dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} }$

उत्तर: ${\text{I = }}\int_{\text{2}}^{\text{3}} {\dfrac{{{\text{x dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}}$

\begin{align}{{\text{x}}^{\text{2}}}{\text{ + 1 = t}} \hfill \\ {\text{2x dx = dt}} \hfill \\ \end{align}

${\text{x dx = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{dt}}$

\begin{align}{\text{x = 3, t = 10 ; x = 2, t = 5}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{5}}^{{\text{10}}} {\dfrac{{{\text{dt}}}}{{\text{t}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[logt]}}_{\text{5}}^{{\text{10}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[log10 - log5]}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log}}\dfrac{{{\text{10}}}}{{\text{5}}} \hfill \\ {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log2}} \hfill \\ \end{align}

14. $\mathbf{\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{2x + 3}}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{2x + 3}}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{{\text{2x}}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}{\text{ + }}\dfrac{{\text{3}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{10x}}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{dx + }}\dfrac{{\text{3}}}{{\text{5}}}\int_{\text{0}}^{\text{1}}{\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\text{1}}}{{\sqrt {\text{5}} }}} \right)}^{\text{2}}}}}} {\text{dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{10x}}}}{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{dx}} \hfill \\{\text{t = 5}}{{\text{x}}^{\text{2}}}{\text{ + 1}} \hfill \ {\text{dt = 10 x dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{dt}}}}{{\text{t}}}} {\text{ = log t}} \hfill \\{\text{ = log}}\left( {{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right) \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{5}}}\left[ {{\text{log}}\left( {{\text{5}}{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right]_{\text{0}}^{\text{1}}{\text{ + }}\dfrac{{\text{3}}}{{\text{5}}}{\text{ \times }}\dfrac{{\text{1}}}{{\dfrac{{\text{1}}}{{\sqrt {\text{5}} }}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\text{x}}}{{\dfrac{{\text{1}}}{{\sqrt {\text{5}} }}}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}{\text{(log6 - log1) + }}\dfrac{{\text{3}}}{{\sqrt {\text{5}} }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{5}} {\text{ - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{0}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}{\text{log6 + }}\dfrac{{\text{3}}}{{\sqrt {\text{5}} }}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{5}} \hfill \\ \end{align}

15. $\mathbf{\int_{\text{0}}^{\text{1}} {\text{x}} {{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\text{x}} {{\text{e}}^{{{\text{x}}^{\text{2}}}}}{\text{dx}}$

\begin{align}{{\text{x}}^{\text{2}}}{\text{ = t, 2x dx = dt}} \hfill \\{\text{xdx = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{dt}} \hfill \\{\text{x = 1, t = 1 ; x = 0, t = 0}} \hfill \\ \end{align}

\begin{align}{\text{I = }}\int_{\text{0}}^{\text{1}} {\dfrac{{\text{1}}}{{\text{2}}}} {\text{ }}{{\text{e}}^{\text{t}}}{\text{ dt = }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\text{1}} {{{\text{e}}^{\text{t}}}} {\text{ dt}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{{\text{e}}^{\text{t}}}} \right]_{\text{0}}^{\text{1}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left( {{{\text{e}}^{\text{1}}}{\text{ - }}{{\text{e}}^{\text{0}}}} \right) \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(e - 1)}} \hfill \\ \end{align}

16.$\mathbf{\int_{\text{1}}^{\text{2}}{\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{1}}^{\text{2}} {\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}}$

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{2}} {\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}}\hfill\\\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}{\text{ = 5 - }}\dfrac{{{\text{20x + 15}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}} \hfill \\\dfrac{{{\text{20x + 15}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}{\text{ = }}\dfrac{{{\text{20x + 15}}}}{{{\text{(x + 3)(x + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{{\text{(x + 3)}}}}{\text{ + }}\dfrac{{\text{B}}}{{{\text{(x + 1)}}}} \hfill \\ {\text{20x + 15 = A(x + 1) + B(x + 3)}} \hfill \\{\text{x + 3 = 0, x = - 3}} \hfill \\{\text{ - 60 + 15 = A( - 2)}} \hfill \\{\text{45 = 2A}} \hfill \\{\text{A = }}\dfrac{{{\text{45}}}}{{\text{2}}} \hfill \\{\text{x + 1 = 0, x = - 1}} \hfill \\{\text{ - 20 + 15 = 2B}} \hfill \\{\text{ - 5 = 2B}} \hfill \\ \end{align}

\begin{align}{\text{B = }}\dfrac{{{\text{ - 5}}}}{{\text{2}}} \hfill \\\dfrac{{{\text{20x + 15}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}{\text{ = }}\dfrac{{{\text{45}}}}{{{\text{2(x + 3)}}}}{\text{ - }}\dfrac{{\text{5}}}{{{\text{2(x + 1)}}}} \hfill \\ \end{align}

${{\text{I}}_{\text{1}}}{\text{ = }}\int {\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}} {\text{dx = }}\int {\left[ {{\text{5 - }}\dfrac{{{\text{45}}}}{{{\text{2(x + 3)}}}}{\text{ + }}\dfrac{{\text{5}}}{{{\text{2(x + 1)}}}}} \right]} {\text{dx}}$

${\text{ = 5x - }}\dfrac{{{\text{45}}}}{{\text{2}}}{\text{log(x + 3) + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log(x + 1)}}$

\begin{align}{\text{I = }}\int_{\text{1}}^{\text{2}} {\dfrac{{{\text{5}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4x + 3}}}}} {\text{dx = }}\left[ {{\text{5x - }}\dfrac{{{\text{45}}}}{{\text{2}}}{\text{log(x + 3) + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{log(x + 1)}}} \right]_{\text{1}}^{\text{2}} \hfill \\{\text{ = 5 - }}\left\{ {\dfrac{{{\text{45}}}}{{\text{2}}}{\text{log5 - log4}}} \right\}{\text{ + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{(log3 - log2)}} \hfill \\{\text{ = 5 - }}\dfrac{{\text{5}}}{{\text{2}}}\left( {{\text{9log}}\dfrac{{\text{5}}}{{\text{4}}}{\text{ - log}}\dfrac{{\text{3}}}{{\text{2}}}} \right){\text{n\backslash endaligned}} \hfill \\ \end{align}

17.$\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x + }}{{\text{x}}^{\text{3}}}{\text{ + 2}}} \right)} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x + }}{{\text{x}}^{\text{3}}}{\text{ + 2}}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\left[ {{\text{2tan}}\dfrac{{{\pi }}}{{\text{4}}}{\text{ + }}\dfrac{{{{\text{x}}^{\text{4}}}}}{{\text{4}}}{\text{ + 2x}}} \right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\{\text{ = }}\left[ {{\text{2tan}}\dfrac{{{\pi }}}{{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{4}}}{{\left( {\dfrac{{{\pi }}}{{\text{4}}}} \right)}^{\text{4}}}{\text{ + 2}}\left( {\dfrac{{{\pi }}}{{\text{4}}}} \right){\text{ - 0}}} \right] \hfill \\{\text{ = }}\;{\text{2 + }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ + }}\dfrac{{{{{\pi }}^{\text{4}}}}}{{{\text{1024}}}} \hfill \\ \end{align}

18. $\mathbf{\int_{\text{0}}^{{\pi }} {\left( {{\text{si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}} \right)} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{{\pi }} {\left( {{\text{si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}{\text{ - co}}{{\text{s}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}} \right)} {\text{dx}}$

\begin{align}{\text{ = - }}\int_{\text{0}}^{{\pi }} {{\text{cos}}} {\text{x dx}} \hfill \\{\text{ = [sinx]}}_{\text{0}}^{{\pi }} \hfill \\ {\text{ = [0 - 0] = 0}} \hfill \\ \end{align}

19. $\mathbf{\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{6x + 3}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{6x + 3}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{6x}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx + }}\int_{\text{0}}^{\text{2}} {\dfrac{{\text{3}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx}} \hfill \\{\text{ = 3}}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx + 3}}\int_{\text{0}}^{\text{2}} {\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx}} \hfill \\ \end{align}

${{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{2x}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 4}}}}} {\text{dx}}$

\begin{align}{\text{t = }}{{\text{x}}^{\text{2}}}{\text{ + 4}} \hfill \\{\text{2x dx = dt}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{dt}}}}{{\text{t}}}} \hfill \\{\text{ = logt = log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right) \hfill \\{\text{I = 3}}\left[ {{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)} \right]_{\text{0}}^{\text{2}}{\text{ + 3}}\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{-1}}}}\dfrac{{\text{x}}}{{\text{2}}}} \right]_{\text{0}}^{\text{2}} \hfill \\ \end{align}

\begin{align}{\text{ = 3[log8 - log4] + }}\dfrac{{\text{3}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{0}}} \right] \hfill \\{\text{ = 3log2 + }}\dfrac{{{{3\pi }}}}{{\text{8}}} \hfill \\ \end{align}

20. $\mathbf{\int_{\text{0}}^{\text{1}} {\left( {{\text{x}}{{\text{e}}^{\text{x}}}{\text{ + sin}}\dfrac{{{{\pi x}}}}{{\text{4}}}} \right)} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\left( {{\text{x}}{{\text{e}}^{\text{x}}}{\text{ + sin}}\dfrac{{{{\pi x}}}}{{\text{4}}}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{1}} {\text{x}} {{\text{e}}^{\text{x}}}{\text{dx + }}\int_{\text{0}}^{\text{1}} {{\text{sin}}} \dfrac{{{{\pi x}}}}{{\text{4}}}{\text{dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {\text{x}} {{\text{e}}^{\text{x}}}{\text{dx = x}}\int {{{\text{e}}^{\text{x}}}} {\text{dx - }}\int {\left( {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}}\int {{{\text{e}}^{\text{x}}}} {\text{dx}}} \right)} \hfill \\{\text{ = x}}{{\text{e}}^{\text{x}}}{\text{ - }}\int {\text{1}} \left( {{{\text{e}}^{\text{x}}}} \right){\text{dx}} \hfill \\{\text{I = }}\left[ {{\text{x}}{{\text{e}}^{\text{x}}}} \right]_{\text{0}}^{\text{1}}{\text{ - }}\int_{\text{0}}^{\text{1}} {{{\text{e}}^{\text{x}}}} {\text{dx - }}\dfrac{{\text{4}}}{{{\pi }}}\left[ {{\text{cos}}\dfrac{{{{\pi x}}}}{{\text{4}}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\left[ {{\text{x}}{{\text{e}}^{\text{x}}}} \right]_{\text{0}}^{\text{1}}{\text{ - }}\left[ {{{\text{e}}^{\text{x}}}} \right]_{\text{0}}^{\text{1}}{\text{ - }}\dfrac{{\text{4}}}{{{\pi }}}\left[ {{\text{cos}}\dfrac{{{{\pi x}}}}{{\text{4}}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = e - }}\left( {{{\text{e}}^{\text{1}}}{\text{ - }}{{\text{e}}^{\text{0}}}} \right){\text{ - }}\dfrac{{\text{4}}}{{{\pi }}}\left( {\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{ - 1}}} \right) \hfill \\{\text{ = e - e + 1 - }}\dfrac{{\text{4}}}{{{{\pi }}\sqrt {\text{2}} }}{\text{ + }}\dfrac{{\text{4}}}{{{\pi }}} \hfill \\{\text{ = 1 + }}\dfrac{{\text{4}}}{{{\pi }}}{\text{ - }}\dfrac{{{\text{2}}\sqrt {\text{2}} }}{{{\pi }}} \hfill \\ \end{align}

### प्रश्न 21 एवं 22 मे सही उत्तर का चयन कीजिए:

21. $\mathbf{\int_{\text{1}}^{\sqrt {\text{3}} } {\dfrac{{{\text{dx}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} }$ बराबर है।

1. $\mathbf{\dfrac{{{\pi }}}{{\text{3}}}}$

2. $\mathbf{\dfrac{{{{2\pi }}}}{{\text{3}}}}$

3. $\mathbf{\dfrac{{{\pi }}}{{\text{6}}}}$

4. $\mathbf{\dfrac{{{\pi }}}{{{\text{12}}}}}$

उत्तर: $\int_{\text{1}}^{\sqrt {\text{3}} } {\dfrac{{{\text{dx}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}}$

\begin{align}{\text{ = }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{1}}^{\sqrt {\text{3}} } \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{3}}}{\text{ - }}\dfrac{{{\pi }}}{{\text{4}}}{\text{ = }}\dfrac{{{\pi }}}{{{\text{12}}}} \hfill \\ \end{align}

अतः विकल्प (d) सही है।

22.$\mathbf{\int_{\text{0}}^{\dfrac{{\text{2}}}{{\text{3}}}}{\dfrac{{{\text{dx}}}}{{{\text{4 + 9}}{{\text{x}}^{\text{2}}}}}} {$ बराबर है।

1. $\dfrac{{{\pi }}}{{\text{6}}}$

2. $\dfrac{{{\pi }}}{{{\text{12}}}}$

3. $\dfrac{{{\pi }}}{{{\text{24}}}}$

4. $\dfrac{{{\pi }}}{{\text{4}}}$

उत्तर: $\int_{\text{0}}^{\dfrac{{\text{2}}}{{\text{3}}}} {\dfrac{{{\text{dx}}}}{{{\text{4 + 9}}{{\text{x}}^{\text{2}}}}}}$

\begin{align}{\text{I = }}\int_{\text{0}}^{\dfrac{{\text{2}}}{{\text{3}}}} {\dfrac{{{\text{dx}}}}{{{\text{4 + 9}}{{\text{x}}^{\text{2}}}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{9}}}\int_{\text{0}}^{\dfrac{{\text{2}}}{{\text{3}}}}{\dfrac{{{\text{dx}}}}{{{{\left( {\dfrac{{\text{2}}}{{\text{3}}}} \right)}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{6}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{{{\text{3x}}}}{{\text{2}}}} \right)} \right]_{\text{0}}^{\dfrac{{\text{2}}}{{\text{3}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{6}}}{\text{ \times }}\dfrac{{{\pi }}}{{\text{4}}}{\text{ = }}\dfrac{{{\pi }}}{{{\text{24}}}} \hfill \\ \end{align}

अतः विकल्प (c) सही है।

### प्रश्नावली 7.10

1 से 8 तक के प्रश्नों समाकलनों का मान प्रतिस्थापन का उपयोग करते हुए ज्ञात कीजिए:

1. $\mathbf{\int_{\text{0}}^{\text{1}} {\dfrac{{\text{x}}}{{{{\text{x}}^{\text{2}}}{\text{ + 1}}}}} {\text{dx}}}$

उत्तर: माना ${{\text{x}}^{\text{2}}}{\text{ + 1 = t}}$

\begin{align}{\text{2x + dx = dt}} \hfill \\{\text{x = 1, t = 2}} \hfill \\{\text{x = 0, t = 1}} \hfill \\{\text{I = }}\int_{\text{1}}^{\text{2}} {\dfrac{{{\text{dt}}}}{{{\text{2t}}}}} {\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[logt]}}_{\text{1}}^{\text{2}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[log2 - log1] = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{log2}} \hfill \\ \end{align}

2. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\sqrt {{\text{sin}}\emptyset } } {\text{ co}}{{\text{s}}^{\text{2}}}\emptyset {\text{ d}}\emptyset }$

उत्तर: ${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\sqrt {{\text{sin}}\emptyset } } {\left( {{\text{1 - si}}{{\text{n}}^{\text{2}}}} \right)^{\text{2}}}{\text{cos}}\emptyset {\text{d}}\emptyset$

माना ${\text{sin}}\emptyset \;{\text{ = t}}$

${\text{cos}}\emptyset {\text{ d}}\emptyset \;{\text{ = dt}}$

\begin{align}\emptyset \;{\text{ = 0, t = 0}} \hfill \\\emptyset \;{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{, t = 1}} \hfill \\{\text{I = }}\int_{\text{0}}^{\text{1}} {\sqrt {\text{t}} } {{\text{(1 - t)}}^{{{\text{2}}^{\text{2}}}}}{\text{dt = }}\int_{\text{0}}^{\text{1}} {\sqrt {\text{t}} } \left( {{\text{1 - 2}}{{\text{t}}^{\text{2}}}{\text{ + }}{{\text{t}}^{\text{1}}}} \right){\text{dt}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{1}} {\left( {{{\text{t}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{+}}{{\text{t}}^{\dfrac{{\text{9}}}{{\text{2}}}}}{\text{ - 2}}{{\text{t}}^{\dfrac{{\text{5}}}{{\text{2}}}}}} \right)} {\text{dt = }}\left[ {\dfrac{{\text{2}}}{{\text{3}}}{{\text{t}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ + }}\dfrac{{\text{2}}}{{{\text{11}}}}{{\text{t}}^{\dfrac{{{\text{11}}}}{{\text{2}}}}}{\text{ - }}\dfrac{{\text{4}}}{{\text{7}}}{{\text{t}}^{\dfrac{{\text{7}}}{{\text{2}}}}}}\right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\left( {{{\text{1}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ - 0}}} \right){\text{ + }}\dfrac{{\text{2}}}{{{\text{11}}}}\left( {{{\text{1}}^{\dfrac{{{\text{11}}}}{{\text{2}}}}}{\text{ -0}}} \right){\text{ - }}\dfrac{{\text{4}}}{{\text{7}}}\left( {{{\text{1}}^{\dfrac{{\text{7}}}{{\text{2}}}}}{\text{ - 0}}} \right) \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ + }}\dfrac{{\text{2}}}{{{\text{11}}}}{\text{ - }}\dfrac{{\text{4}}}{{\text{7}}}{\text{ = }}\dfrac{{{\text{154 + 42 - 132}}}}{{{\text{3 \times 11 \times 7}}}} \hfill \\ {\text{ = }}\dfrac{{{\text{64}}}}{{{\text{231}}}} \hfill \\ \end{align}

3. $\int_{\text{0}}^{\text{1}} {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{2x}}}}{{{\text{1 + }}{{\text{x}}^{\text{2}}}}}} \right){\text{dx}}$

उत्तर: माना ${\text{x = tan\theta }}$

\begin{align}{\text{dx = se}}{{\text{c}}^{\text{2}}}{\text{\theta d\theta }} \hfill \\{\text{x = 0, \theta = 0}} \hfill \\{\text{x = 1, \theta = }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\{\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{2tan\theta }}}}{{{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{\theta }}}}} \right){\text{se}}{{\text{c}}^{\text{2}}}{\text{\theta d\theta }} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\text{2}} {\text{\theta se}}{{\text{c}}^{\text{2}}}{\text{\theta d\theta }} \hfill \\{\text{ = [2\theta tan\theta - logsec\theta ]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - log2}} \hfill \\ \end{align}

4. $\mathbf{\int_{\text{0}}^{\text{2}} {\text{x}} \sqrt {{\text{x + 2}}} {\text{dx }}\left( {{\text{x + 2 = }}{{\text{t}}^{\text{2}}}} \right.{\text{)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{\text{2}} {{\text{x}}\sqrt {{\text{x + 2}}} } \;{\text{dx}}$

माना ${\text{x + 2 = }}{{\text{t}}^{\text{2}}}$

\begin{align}{\text{dx = 2t dt}} \hfill \\{\text{x = 0, t = }}\sqrt {\text{2}} \hfill \\{\text{x = 2, }}{{\text{t}}^{\text{2}}}{\text{ = 4;t = 2}} \hfill \\{\text{I = }}\int_{\sqrt {\text{2}} }^{\text{2}} {\left( {{{\text{t}}^{\text{2}}}{\text{ - 2}}} \right)} {\text{ \times t2t}}{\text{.dt = 2}}\int_{\sqrt {\text{2}} }^{\text{2}} {\left( {{{\text{t}}^{\text{4}}}{\text{ - 2}}{{\text{t}}^{\text{2}}}} \right)} {\text{dt [(x + 2)}}\;{\text{ = }}\;{{\text{t}}^{\text{2}}}{\text{ ; x = }}{{\text{t}}^{\text{2}}}{\text{ - 2]}} \hfill \\{\text{ = 2}}\left[ {\dfrac{{{{\text{t}}^{\text{5}}}}}{{\text{5}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{t}}^{\text{3}}}} \right]_{\sqrt {\text{2}} }^{\text{2}} \hfill \\{\text{ = 2}}\left[ {\left( {\dfrac{{{{\text{2}}^{\text{5}}}}}{{\text{5}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{{{\text{(2)}}}^{\text{3}}}} \right){\text{ - }}\left( {\dfrac{{{{{\text{(}}\sqrt {\text{2}} {\text{)}}}^{\text{5}}}}}{{\text{5}}}{\text{ - }}\dfrac{{\text{2}}}{{\text{3}}}{{{\text{(}}\sqrt {\text{2}} {\text{)}}}^{\text{3}}}} \right)} \right] \hfill \\{\text{ = 2}}\left[ {\dfrac{{{\text{32}}}}{{\text{5}}}{\text{ - }}\dfrac{{{\text{16}}}}{{\text{3}}}} \right]{\text{ - 2}}\left( {\dfrac{{{\text{4}}\sqrt {\text{2}} }}{{\text{5}}}{\text{ - }}\dfrac{{{\text{4}}\sqrt {\text{2}} }}{{\text{3}}}} \right){\text{ = }}\dfrac{{{\text{64}}}}{{\text{5}}}{\text{ - }}\dfrac{{{\text{32}}}}{{\text{3}}}{\text{ - }}\dfrac{{{\text{8}}\sqrt {\text{2}} }}{{\text{5}}}{\text{ + }}\dfrac{{{\text{8}}\sqrt {\text{2}} }}{{\text{3}}} \hfill \\{\text{ = }}\dfrac{{{\text{192 - 160}}}}{{{\text{15}}}}{\text{ + }}\dfrac{{{\text{8}}\sqrt {\text{2}} {\text{(5 - 3)}}}}{{{\text{15}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{15}}}}{\text{[32 + 16}}\sqrt {\text{2}} {\text{] = }}\dfrac{{{\text{16}}}}{{{\text{15}}}}{\text{(2 + }}\sqrt {\text{2}} {\text{)}} \hfill \\ \end{align}

5. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sinx}}}}{{{\text{1 + co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} {\text{dx}}}$

उत्तर: ${\text{I}}\,{\text{ = }}\;\int\limits_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sinx}}}}{{{\text{1 + co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} {\text{dx}}$

माना ${\text{cosx}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{sinx}}\;{\text{dx = dt}} \hfill \\{\text{x = 0, t = cos 0 = 1}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{,t = cos}}\dfrac{{{\pi }}}{{\text{2}}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

\begin{align}{\text{I = }}\int_{\text{1}}^{\text{0}} {\dfrac{{{\text{ - dt}}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} {\text{ = - }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]_{\text{1}}^{\text{0}}{\text{ = - }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{0 - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1}}} \right] \hfill \\{\text{ = - }}\left[ {{\text{ - }}\dfrac{{{\pi }}}{{\text{4}}}} \right]{\text{ = }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

6. $\mathbf{\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{dx}}}}{{{\text{x + 4 - }}{{\text{x}}^{\text{2}}}}}} }$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{dx}}}}{{{\text{x + 4 - }}{{\text{x}}^{\text{2}}}}}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{dx}}}}{{{\text{4 - }}\left( {{{\text{x}}^{\text{2}}}{\text{ - x + }}\dfrac{{\text{1}}}{{\text{4}}}} \right){\text{ + }}\dfrac{{\text{1}}}{{\text{4}}}}}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{2}} {\dfrac{{{\text{dx}}}}{{\dfrac{{{\text{17}}}}{{\text{4}}}{\text{ - }}{{\left( {{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}^{\text{2}}}}}} \;\,\;\;\;\,\;\,\left[ {\int {\dfrac{{\text{1}}}{{{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}}}} {\text{dx = }}\dfrac{{\text{1}}}{{{\text{2a}}}}{\text{log}}\dfrac{{{\text{a + x}}}}{{{\text{a - x}}}}{\text{ + c}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{2}}\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}}}{\text{log}}\left[ {\dfrac{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ + }}\left( {{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ - }}\left( {{\text{x - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}} \right]_{\text{0}}^{\text{2}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}\left[ {{\text{log}}\left( {\dfrac{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ + }}\left( {{\text{2 - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ - }}\left( {{\text{2 - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}} \right)} \right]{\text{ - log}}\left( {\dfrac{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ + }}\left( {{\text{0 - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ - }}\left( {{\text{0 - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}} \right) \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}\left[ {{\text{log}}\left( {\dfrac{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ - }}\dfrac{{\text{3}}}{{\text{2}}}}}} \right){\text{ - log}}\left( {\dfrac{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ - }}\dfrac{{\text{3}}}{{\text{2}}}}}{{\dfrac{{\sqrt {{\text{17}}} }}{{\text{2}}}{\text{ + }}\dfrac{{\text{3}}}{{\text{2}}}}}} \right)} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}\left[ {{\text{log}}\left( {\dfrac{{\sqrt {{\text{17}}} {\text{ + 3}}}}{{\sqrt {{\text{17}}} {\text{ - 3}}}}{\text{ \times }}\dfrac{{\sqrt {{\text{17}}} {\text{ + 3}}}}{{\sqrt {{\text{17}}} {\text{ - 3}}}}} \right)} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\left( {\dfrac{{{\text{20 + 4}}\sqrt {{\text{17}}} }}{{{\text{20 - 4}}\sqrt {{\text{17}}} }}} \right) \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\left( {\dfrac{{{\text{5 + }}\sqrt {{\text{17}}} }}{{{\text{5 - }}\sqrt {{\text{17}}} }}} \right){\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\left( {\dfrac{{{\text{5 + }}\sqrt {{\text{17}}} }}{{{\text{5 - }}\sqrt {{\text{17}}} }}} \right)\left( {\dfrac{{{\text{5 + }}\sqrt {{\text{17}}} }}{{{\text{5 - }}\sqrt {{\text{17}}} }}} \right) \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\dfrac{{{{{\text{(5 + }}\sqrt {{\text{17}}} {\text{)}}}^{\text{2}}}}}{{{\text{25 - 17}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\dfrac{{{\text{25 + 17 - 10}}\sqrt {{\text{17}}} }}{{\text{8}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\dfrac{{{\text{42 - 10}}\sqrt {{\text{17}}} }}{{\text{8}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\sqrt {{\text{17}}} }}{\text{log}}\dfrac{{{\text{21 - 5}}\sqrt {{\text{17}}} }}{{\text{5}}} \hfill \\ \end{align}

7. $\mathbf{\int_{{\text{ - 2}}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 2x + 5}}}}} }$

उत्तर: ${\text{I = }}\int_{{\text{ - 2}}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 2x + 5}}}}}$

\begin{align}{\text{ = }}\int_{{\text{ - 1}}}^{\text{1}} {\dfrac{{{\text{dx}}}}{{{{{\text{(x + 1)}}}^{\text{2}}}{\text{ + 4}}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{x + 1}}}}{{\text{2}}}} \right]_{{\text{ - 1}}}^{\text{1}}\;\;\;\,\;\;\;\;\left[ {\int {\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}}}} {\text{dx = }}\dfrac{{\text{1}}}{{\text{a}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right] \hfill \\= \;\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{{\text{1 + 1}}}}{{\text{2}}}{\text{ - tan}}\dfrac{{{\text{1 - 1}}}}{{\text{2}}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ \times }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{8}}} \hfill \\ \end{align}

8. $\mathbf{\int_{\text{1}}^{\text{2}} {\left( {\dfrac{{\text{1}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}} \right)} {{\text{e}}^{{\text{2x}}}}{\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{1}}^{\text{2}} {\left( {\dfrac{{\text{1}}}{{\text{x}}}{\text{ - }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}} \right)} {{\text{e}}^{{\text{2x}}}}{\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{\text{x}}}{\text{dx - }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}{\text{dx}} \hfill \\{\text{ = }}{{\text{I}}_{\text{1}}}{\text{ - }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}{\text{dx}} \hfill \\\end{align}

${{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{\text{x}}}{\text{dx}}$

${\text{ = }}\left[ {\dfrac{{\text{1}}}{{\text{x}}}\dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}}\right]_{\text{1}}^{\text{2}}{\text{ - }}\int_{\text{1}}^{\text{2}} {\left( {{\text{ - }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}} \right)} \dfrac{{{{\text{e}}^{{\text{2x}}}}}}{{\text{2}}}{\text{dx}}$$\dfrac{{\text{1}}}{{\text{x}}}$ को पहला फलन तथा ${{\text{e}}^{{\text{2x}}}}$ दूसरे का फलन लेकर समाकलन करने पर

\begin{align}{\text{ = }}\left( {\dfrac{{\text{1}}}{{\text{2}}}\dfrac{{{{\text{e}}^{\text{4}}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{1}}}\dfrac{{{{\text{e}}^{\text{2}}}}}{{\text{2}}}} \right){\text{ + }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}{\text{dx}} \hfill \\{\text{I = }}\dfrac{{{{\text{e}}^{\text{2}}}}}{{\text{4}}}\left( {{{\text{e}}^{\text{2}}}{\text{ - 2}}} \right){\text{ + }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}{\text{dx - }}\int_{\text{1}}^{\text{2}} {{{\text{e}}^{{\text{2x}}}}} {\text{ \times }}\dfrac{{\text{1}}}{{{\text{2}}{{\text{x}}^{\text{2}}}}}{\text{dx}} \hfill \\{\text{ = }}\dfrac{{{{\text{e}}^{\text{2}}}}}{{\text{4}}}\left( {{{\text{e}}^{\text{2}}}{\text{ - 2}}} \right) \hfill \\ \end{align}

प्रश्न 9 एवं 10 मे सही उत्तर का चयन कीजिए:

9. समाकलन $\mathbf{\int_{\dfrac{{\text{1}}}{{\text{3}}}}^{\text{1}} {\dfrac{{{{\left( {{\text{x - }}{{\text{x}}^{\text{3}}}} \right)}^{\dfrac{{\text{1}}}{{\text{3}}}}}}}{{{{\text{x}}^{\text{4}}}}}} {\text{dx}}}$ का मान है।

(a) $\mathbf{6}$

(b) $\mathbf{0}$

(c) $\mathbf{3}$

(d) $\mathbf{4}$

उत्तर: ${\text{I = }}\int_{\dfrac{{\text{1}}}{{\text{3}}}}^{\text{1}} {\dfrac{{{{\left( {{\text{x - }}{{\text{x}}^{\text{2}}}} \right)}^{\dfrac{{\text{1}}}{{\text{3}}}}}}}{{{{\text{x}}^{\text{4}}}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\dfrac{{\text{1}}}{{\text{3}}}}^{\text{1}} {\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{3}}}}}{{\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)}^{\dfrac{{\text{1}}}{{\text{3}}}}}}}{{{{\text{x}}^{\text{4}}}}}} {\text{dx}} \hfill \\ {\text{1 - }}{{\text{x}}^{\text{2}}}{\text{ = }}{{\text{t}}^{\text{3}}} \hfill \\{{\text{x}}^{\text{2}}}{\text{ = 1 - }}{{\text{t}}^{\text{3}}} \hfill \\{\text{2x dx = 3}}{{\text{t}}^{\text{2}}} \hfill \\ \end{align}

\begin{align}{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{{\text{x}}^{\dfrac{{\text{1}}}{{\text{3}}}}}{{\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)}^{\dfrac{{\text{1}}}{{\text{3}}}}}{\text{( - 2x)}}}}{{{{\text{x}}^{\text{4}}}{\text{x}}}}} {\text{dx}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{{\left( {{\text{1 - }}{{\text{t}}^{\text{2}}}} \right)}^{\dfrac{{\text{1}}}{{\text{3}}}}}}}{{{{\left( {{\text{1 - }}{{\text{t}}^{\text{2}}}} \right)}^{\text{2}}}}}} {\text{3}}{{\text{t}}^{\text{2}}}{\text{ dt}} \hfill \\ \end{align}

$\left. {{\text{ = }}\dfrac{{{\text{ - 3}}}}{{{\text{4(x}}{{\text{)}}^{\text{2}}}}}{\text{ = }}\dfrac{{{\text{ - 3}}}}{{\text{4}}}{{\text{x}}^{\text{2}}}} \right]_{\dfrac{{\text{1}}}{{\text{3}}}}^{\text{1}}$

${\text{ = 6}}$

अतः विकल्प (a) सही है।

10. यदि $\mathbf{{\text{f(x)}}\;{\text{ = }}\;\int\limits_{\text{0}}^{\text{x}} {{\text{t}}\;{\text{sint}}\;{\text{dt}}}$ तब ${{\text{f}}^'}{\text{(x)}}}$

1. $\mathbf{{\text{cosx + sinx}}}$

2. $\mathbf{{\text{x}}\;{\text{sinx}}}$

3. $\mathbf{{\text{x}}\;{\text{cosx}}}$

4. $\mathbf{{\text{sinx + cosx}}}$

उत्तर: ${\text{f(x)}}\;{\text{ = }}\;\int\limits_{\text{0}}^{\text{x}} {{\text{t}}\;{\text{sint}}\;{\text{dt}}}$

\begin{align} {\text{ = t( - cost) - }}\int {\text{1}} {\text{[( - cost)dt]}}_{\text{0}}^{\text{x}} \hfill \\{\text{ = xcosx + sinx}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

### प्रश्नावली 7.11

निश्चित समाकलनों के गुणधर्मों का उपयोग करते हुए 1 से 19 के प्रश्नों के समाकलनों का मान ज्ञात कीजिए:

1. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{xdx}}}$

उत्तर: $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{1 - cos(2x)}}}}{{\text{2}}}} {\text{dx}}$

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(1 + cos2x)}}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{1}} {\text{ dx + }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{cos}}} {\text{2x dx}} \hfill \\= \;\dfrac{{\text{1}}}{{\text{2}}}{\text{[x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{(sin2x)}}} \right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}[\left[ {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 0}}} \right]{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{sin\pi - 0}}} \right]] \hfill \\ = \;\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

2. $mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{\sqrt {{\text{sinx}}} }}{{\sqrt {{\text{sinx}}} {\text{ + }}\sqrt {{\text{cosx}}} }}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{\sqrt {{\text{sinx}}} }}{{\sqrt {{\text{sinx}}} {\text{ + }}\sqrt {{\text{cosx}}} }}} {\text{dx}}$ …………….(i)

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{\sqrt {{\text{sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} }}{{\sqrt {{\text{sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} {\text{ + }}\sqrt {{\text{sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} }}} {\text{dx}}$

${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{\sqrt {{\text{cosx}}} }}{{\sqrt {{\text{cosx}}} {\text{ + }}\sqrt {{\text{sinx}}} }}} {\text{dx}}$ …………………(ii)

(i) +(ii)

\begin{align}{\text{2I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{1 }}} {\text{dx}} \hfill \\{\text{ = [x]}}_{\text{0}}^{\dfrac{{\text{\pi }}}{{\text{2}}}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}\; = \;\dfrac{{{\pi }}}{4} \hfill \\ \end{align}

3. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x}}}}{{{\text{si}}{{\text{n}}^{\dfrac{3}{2}}}{\text{x + co}}{{\text{s}}^{\dfrac{3}{{\text{2}}}}}{\text{x}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x}}}}{{{\text{si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x + co}}{{\text{s}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x}}}}} {\text{dx}}$ ……….(i)

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}{{{\text{si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right){\text{ + co}}{{\text{s}}^{\dfrac{{\text{3}}}{{\text{2}}}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{co}}{{\text{s}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x + si}}{{\text{n}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{x}}}}} {\text{dx}}$ …………(ii)

(i) + (ii)

\begin{align}{\text{2I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{1}} {\text{ dx = [x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\ {\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ = }}\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

4.$\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}}}{{{\text{si}}{{\text{n}}^{\text{5}}}{\text{x + co}}{{\text{s}}^{\text{5}}}{\text{x}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}}}{{{\text{si}}{{\text{n}}^{\text{5}}}{\text{x + co}}{{\text{s}}^{\text{5}}}{\text{x}}}}} {\text{dx}}$ ………………(i)

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{co}}{{\text{s}}^{\text{5}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}{{{\text{si}}{{\text{n}}^{\text{5}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right){\text{ + co}}{{\text{s}}^{\text{5}}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{si}}{{\text{n}}^{\text{5}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{5}}}{\text{x + si}}{{\text{n}}^{\text{5}}}{\text{x}}}}} {\text{dx}}$ ……………………(ii)

(i) + (ii)

\begin{align}{\text{2I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{1}} {\text{ dx = [x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}\; = \;\dfrac{{{\pi }}}{{\text{4}}} \hfill \\ \end{align}

5. $\mathbf{\int_{{\text{ - 5}}}^{\text{5}} {\text{|}} {\text{x + 2|dx}}}$

उत्तर: ${\text{I = }}\int_{{\text{ - 5}}}^{\text{5}} {\text{|}} {\text{x + 2| dx}}$

\begin{align}{\text{ - }}\int_{{\text{ - 5}}}^{{\text{ - 2}}} {{\text{(x + 2)}}} {\text{dx + }}\int_{{\text{ - 2}}}^{\text{5}} {{\text{(x + 2)}}} {\text{dx [}}\because \;\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x)dx}}\; = \;\int_{\text{a}}^{\text{c}} {\text{f}} {\text{(x)dx + }}\int_{\text{c}}^{\text{b}} {\text{f}} {\text{(x)dx}}] \hfill \\{\text{ - }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + 2x}}} \right]_{{\text{ - 5}}}^{{\text{ - 2}}}{\text{ + }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + 2x}}} \right]_{{\text{ - 2}}}^{\text{5}}\quad {\text{[}}\because \;{\text{( - 5, - 2) ; (x + 2)}} \leqslant {\text{0}}\;{\text{,}}\;{\text{( - 2,5) ; (x + 2)}} \geqslant {\text{0]}} \hfill \\{\text{ - }}\left[ {\dfrac{{{\text{ - }}{{\text{2}}^{\text{2}}}}}{{\text{2}}}{\text{ + 2( - 2) - }}\dfrac{{{\text{ - }}{{\text{5}}^{\text{2}}}}}{{\text{2}}}{\text{ - 2( - 5)}}} \right]{\text{ + }}\left[ {\dfrac{{{{\text{5}}^{\text{2}}}}}{{\text{2}}}{\text{ + 2(5) - }}\dfrac{{{\text{ - }}{{\text{2}}^{\text{2}}}}}{{\text{2}}}{\text{ + 2( - 2)}}} \right] \hfill \\\left[ {\left( {\dfrac{{\text{4}}}{{\text{2}}}{\text{ - 4}}} \right){\text{ - }}\left( {\dfrac{{{\text{25}}}}{{\text{2}}}{\text{ - 10}}} \right)} \right]{\text{ + }}\left[ {\left( {\dfrac{{{\text{25}}}}{{\text{2}}}{\text{ + 10}}} \right){\text{ - }}\left( {\dfrac{{\text{4}}}{{\text{2}}}{\text{ - 4}}} \right)} \right] \hfill \\{\text{ - }}\left( {{\text{ - 2 - }}\dfrac{{\text{5}}}{{\text{2}}}} \right){\text{ + }}\left( {\dfrac{{{\text{45}}}}{{\text{2}}}{\text{ + 2}}} \right) \hfill \\{\text{ = 29}} \hfill \\ \end{align}

6. $\mathbf{\int_{\text{2}}^{\text{8}} {\text{|}} {\text{x - 5|dx}}}$

उत्तर: ${\text{I = }}\int_{\text{2}}^{\text{8}} {\text{|}} {\text{x - 5| dx}}$

\begin{align}{\text{ - }}\int_{\text{2}}^{\text{5}} {{\text{(x - 5)}}} {\text{dx + }}\int_{\text{5}}^{\text{8}} {{\text{(x - 5)}}} {\text{dx}} \hfill \\\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 5x}}} \right]_{\text{2}}^{\text{5}}{\text{ + }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 5x}}} \right]_{\text{5}}^{\text{8}}{\text{ = }}\left[ {\left( {\dfrac{{{\text{25}}}}{{\text{2}}}{\text{ - 25}}} \right){\text{ - }}\left( {\dfrac{{\text{4}}}{{\text{2}}}{\text{ - 10}}} \right)} \right]{\text{ + [}}\left( {\dfrac{{{\text{64}}}}{{\text{2}}}{\text{ - 40}}} \right){\text{ - }}\left( {\dfrac{{{\text{25}}}}{{\text{2}}}{\text{ - 25}}} \right){\text{]}} \hfill \\{\text{ = 9}} \hfill \\ \end{align}

7. $\mathbf{\int_{\text{0}}^{\text{1}} {\text{x}} {{\text{(1 - x)}}^{\text{n}}}{\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {\text{x}} {{\text{(1 - x)}}^{\text{n}}}{\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{1}} {{\text{(1 - x)}}} {{\text{[1 - (1 - x)]}}^{\text{n}}}{\text{dx}} \hfill \\ {\text{ = }}\int_{\text{0}}^{\text{1}} {{{{\text{(x)}}}^{\text{n}}}} {{\text{(x)}}^{{\text{n - 1}}}}{\text{dx}} \hfill \\{\text{ = }}\left[ {\dfrac{{{{\text{x}}^{{\text{n + 1}}}}}}{{{\text{n + 1}}}}{\text{ - }}\dfrac{{{{\text{x}}^{{\text{n + 2}}}}}}{{{\text{n + 2}}}}} \right] \hfill \\{\text{ = }}\left( {\dfrac{{\text{1}}}{{{\text{(n + 1)(n + 2)}}}}} \right) \hfill \\ \end{align}

8. $\int_{\text{0}}^{\dfrac{{\text{\pi }}}{{\text{4}}}} {{\text{log}}} {\text{(1 + tanx)dx}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} {\text{(1 + tanx)dx}}$

\begin{align} {\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} \left( {{\text{1 + tan}}\left( {\dfrac{{{\pi }}}{{\text{4}}}{\text{ - x}}} \right)} \right){\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} \left( {{\text{1 + }}\dfrac{{{\text{1 - tanx}}}}{{{\text{1 + tanx}}}}} \right){\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} \left( {\dfrac{{\text{2}}}{{{\text{1 + tanx}}}}} \right){\text{dx}} \hfill \\ {\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} {\text{2dx + }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{log}}} {\text{(1 + tanx)dx}} \hfill \\{\text{I = log2}}\left( {\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\text{1}} {\text{dx}}} \right){\text{ - 1}} \hfill \\ {\text{2I = log2[x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\ {\text{I = }}\dfrac{{{\pi }}}{{\text{8}}}{\text{log2}} \hfill \\ \end{align}

9. $\mathbf{\int_{\text{0}}^{\text{2}} {\text{x}} \sqrt {{\text{2 - x}}} {\text{dx}}}$

उत्तर: ${\text{ = }}\int_{\text{0}}^{\text{2}} {{\text{(2 - x)}}} \sqrt {{\text{2 - (2 - x)}}} {\text{ dx [}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx = }}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(a - x)dx]}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{2}} {{\text{(2 - x)}}} \sqrt {{\text{(x)}}} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{2}} {\text{(}} {\text{2}}\sqrt {{\text{(x)}}} {\text{ - x}}\sqrt {{\text{(x)}}} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{2}} {\left( {{\text{2}}{{\text{x}}^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{-}}{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}} \right)} {\text{dx}} \hfill \\{\text{ = 2x}}_{\text{3}}^{\text{2}}\left[ {{{\text{x}}^{\dfrac{{\text{3}}}{{\text{2}}}}}} \right]_{\text{0}}^{\text{2}}{\text{ - }}\dfrac{{\text{2}}}{{\text{5}}}\left[ {{{\text{x}}^{\dfrac{{{\text{35}}}}{{\text{2}}}}}} \right]_{\text{0}}^{\text{2}} \hfill \\{\text{ = }}\dfrac{{\text{4}}}{{\text{3}}}\left[ {{{\text{2}}^{\dfrac{{\text{3}}}{{\text{2}}}}}{\text{ - 0}}} \right]_{\text{0}}^{\text{2}}{\text{ - }}\dfrac{{\text{2}}}{{\text{5}}}\left[ {{{\text{2}}^{\dfrac{{\text{5}}}{{\text{2}}}}}{\text{ - 0}}} \right]_{\text{0}}^{\text{2}} \hfill \\{\text{ = }}\dfrac{{{\text{16}}\sqrt {{\text{(2)}}} }}{{{\text{15}}}} \hfill \\ \end{align}

10. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{(}} {\text{2 log sinx - log sin2x)dx}}}$

उत्तर: ${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(2logsinx - log2sinxcosx)}}} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(2logsinx - log2 - logsinx - logcosx)}}} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {{\text{2logsinxdx - }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log}}} {\text{2dx - }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {{\text{logcos}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} \right)} {\text{dx}}} \right.}$

\begin{align}{\text{ = log2[x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\ {\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2}} \hfill \\ \end{align}

11. $\int_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{2}}}} {\text{x dx}}$

उत्तर: ${\text{f(x) = si}}{{\text{n}}^{\text{2}}}{\text{x}}$

\begin{align}{\text{f( - x) = [sin( - x)}}{{\text{]}}^{\text{2}}}{\text{ = ( - sinx}}{{\text{)}}^{\text{2}}}{\text{ = si}}{{\text{n}}^{\text{2}}}{\text{x = f(x)}} \hfill \\{\text{2}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{2}}}} {\text{xdx = 2}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{1 - cos2x}}}}{{\text{2}}}} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(1 - cos2x)}}} {\text{dx = }}\left[ {{\text{x - }}\dfrac{{{\text{sin2x}}}}{{\text{2}}}} \right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 0 = }}\dfrac{{{\pi }}}{{\text{2}}} \hfill \\ \end{align}

12. $\mathbf{\int_{\text{0}}^{{\pi }} {\dfrac{{\text{x}}}{{{\text{1 + sinx}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{{\pi }} {\dfrac{{\text{x}}}{{{\text{1 + sinx}}}}} {\text{dx}}$ ……….(i)

${\text{I = }}\int_{\text{0}}^{{\pi }} {\dfrac{{{{\pi - x}}}}{{{{1 + sin(\pi - x)}}}}} {\text{dx}}$

${\text{I = }}\int_{\text{0}}^{{\pi }} {\dfrac{{{{\pi - x}}}}{{{\text{1 + sinx}}}}} {\text{dx}}$ ……………….(ii)

(i) + (ii)

\begin{align}{{2I = \pi }}\int_{\text{0}}^{{\pi }} {\dfrac{{\text{1}}}{{{\text{1 + sinx}}}}} {\text{dx}} \hfill \\{{2I = \pi }}\int_{\text{0}}^{{\pi }} {\dfrac{{{\text{1 - sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} {\text{dx}} \hfill \\{{2I = \pi }}\int_{\text{0}}^{{\pi }} {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - secx tanx}}} \right)} {\text{dx}} \hfill \\ \end{align}

\begin{align} {\text{2I = [ tanx - secx)]}}_{\text{0}}^{\dfrac{{\text{\pi }}}{{\text{2}}}}{{ = 2\pi }} \hfill \\{{I = \pi }} \hfill \\ \end{align}

13. $\int_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{7}}}} {\text{x dx}}$

उत्तर: ${\text{I = }}\int_{\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{7}}}} {\text{x dx}}$

\begin{align}{\text{f(x) = si}}{{\text{n}}^{\text{7}}}{\text{x}} \hfill \\{\text{f( - x) = [sin( - x)}}{{\text{]}}^{\text{7}}}{\text{ = ( - sinx}}{{\text{)}}^{\text{7}}}{\text{ = - si}}{{\text{n}}^{\text{7}}}{\text{x = f( - x)}} \hfill \\\int_{{{ - \pi }}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{7}}}} {\text{x dx = 0}}\quad \;\;\;\;\,[\int_{{\text{ - a}}}^{\text{a}} {\text{f}} {\text{(x)dx}}\;{\text{ = 0]}} \hfill \\ \end{align}

14. $\mathbf{\int_{\text{0}}^{{{2\pi }}} {\left( {{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}} \right)} {\text{dx}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{{{2\pi }}} {{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}\;{\text{dx}}}$

${\text{f(x)}}\;{\text{ = }}\;{\text{si}}{{\text{n}}^{\text{5}}}{\text{x}}$

\begin{align}{\text{f( - x) = [cos( - x)}}{{\text{]}}^{\text{7}}}{\text{ = (cosx}}{{\text{)}}^{\text{7}}}{\text{ = co}}{{\text{s}}^{\text{7}}}{\text{x = f(x)}} \hfill \\{\text{I = }}\int_{{0}}^{{{2\pi }}} {\left( {{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}} \right)} {\text{dx}} \hfill \\{\text{ = 2}}\int_{\text{0}}^{{\pi }} {\left( {{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}} \right)} {\text{dx}} \hfill \\{\text{g(x) = co}}{{\text{s}}^{\text{5}}}{\text{x}} \hfill \\{{g(\pi - x) = co}}{{\text{s}}^{\text{5}}}{{(\pi - x) = - co}}{{\text{s}}^{\text{5}}}{\text{x = - g(x)}} \hfill \\{\text{I = 0}} \hfill \\\therefore \;\int_{\text{0}}^{{{2\pi }}} {\left( {{\text{co}}{{\text{s}}^{\text{5}}}{\text{x}}} \right)} {\text{dx = 0}} \hfill \\ \end{align}

15. $\mathbf{\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sinx - cosx}}}}{{{\text{1 + sinxcosx}}}}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sinx - cosx}}}}{{{\text{1 + sinxcosx}}}}} {\text{dx}}$ ……………..(i)

${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right){\text{ - cos}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}{{{\text{1 + sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right){\text{cos}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}} {\text{dx}}$

${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{cosx - sinx}}}}{{{\text{1 + cosxsinx}}}}} {\text{dx}}$ ……………….(ii)

(i) + (ii)

${\text{2I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{{\text{sinx - cosx + cosx - sinx}}}}{{{\text{1 + cosxsinx}}}}} {\text{dx}}$

${\text{2I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\dfrac{{\text{0}}}{{{\text{1 + cosxsinx}}}}} {\text{dx}}$

${\text{I = 0}}$

16. $\mathbf{\int_{\text{0}}^{{\pi }} {{\text{log}}} {\text{(1 + cosx)dx}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{log(1 + cosx)dx}}}$ ………..(i)

${\text{I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{log(1 + cos(\pi - x))dx}}}$

${\text{ = }}\int_{\text{0}}^{{\pi }} {{\text{log}}} {\text{(1 - cosx)dx}}$ …………….(ii)

\begin{align}{\text{2I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{[log(1 + cosx) + log(1 - cosx]dx}}} \hfill \\{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{[log(1 - co}}{{\text{s}}^{\text{2}}}{\text{x)}}} \hfill \\{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{[log(si}}{{\text{n}}^{\text{2}}}{\text{x)}}} \hfill \\{\text{2I}}\;{\text{ = }}\;{\text{2}}\int\limits_{\text{0}}^{{\pi }} {{\text{log(sinx)dx}}} \hfill \\{\text{I}}\;{\text{ = }}\;\int\limits_{\text{0}}^{{\pi }} {{\text{log(sinx)dx}}} \;{\text{ = }}\;{\text{2}}\int\limits_{\text{0}}^{{{\pi /2}}} {{\text{log(sinx)dx}}} \;{\text{ = }}\;{\text{2}}{{\text{I}}_{\text{1}}} \hfill \\{\text{[}}\int\limits_{\text{0}}^{{\text{2a}}} {{\text{f(x)dx}}\;{\text{ = }}\;{\text{2}}\int\limits_{\text{0}}^{\text{a}}{{\text{f(x)}}\;{\text{dx}}\;{\text{,}}\;{\text{f(2a - x)}}\;{\text{ = }}\;{\text{f(x)}}} } \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log}}} \left( {{\text{sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)} \right){\text{dx}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log}}} {\text{(cosx)dx}} \hfill \\{\text{2}}{{\text{I}}_1}\;{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log}}} {\text{(sinx)dx + }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log}}} {\text{(cosx)dx}} \hfill \\{\text{2}}{{\text{I}}_1}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(log(}}} {\text{sinx) + log(cosx))dx}} \hfill \\{\text{2}}{{\text{I}}_1}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(log(}}} {\text{sinx cosx)dx}} \hfill \\ \end{align}

\begin{align} {\text{2}}{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{log(}}\dfrac{{{\text{sin2x}}}}{{\text{2}}}{\text{)dx}}} \hfill \\ {\text{2}}{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(log(sin2x)dx - }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(log2) }}} {\text{dx}}} \hfill \\ \end{align}

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {({\text{log(sin2x)dx - log2[x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}})}$

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(log(sin2x)dx - }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2 = }}{{\text{I}}_{\text{2}}}{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2}}}$

\begin{align}{{\text{I}}_{\text{2}}}{\text{ = }}\int_{\text{0}}^{\dfrac{{\text{\pi }}}{{\text{2}}}} {{\text{(log(}}} {\text{sin2x)dx}} \hfill \\{\text{2x = t}} \hfill \\{\text{2}}{\text{.dx = dt}} \hfill \\ {\text{x = 0, t = 0}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{2}}}{{, t = \pi }} \hfill \\ \end{align}

\begin{align}{{\text{I}}_{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}\int\limits_{\text{0}}^{{\pi }} {{\text{log(sint)dt}}} \hfill \\{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{ \times 2}}\int\limits_{\text{0}}^{{\pi }} {{\text{log(sint)dt}}} \hfill \\{\text{ = }}\;\int\limits_{\text{0}}^{{{\pi /2}}} {{\text{log(sinx)dx}}} \;{\text{ = }}\;{{\text{I}}_{\text{1}}} \hfill \\{\text{\backslash }}\;{\text{2}}{{\text{I}}_{\text{1}}}\;{\text{ = }}\;{{\text{I}}_{\text{1}}}{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2}} \hfill \\{{\text{I}}_{\text{1}}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2}} \hfill \\ {\text{I}}\;{\text{ = }}\;{\text{2}}{{\text{I}}_{\text{1}}} \hfill \\{\text{I}}\;{\text{ = }}\;{\text{2( - }}\dfrac{{{\pi }}}{{\text{2}}}{\text{log2)}}\;{\text{ = }}\;{{ - \pi log2}} \hfill \\ \end{align}

17. $\mathbf{\int_{\text{0}}^{\text{a}} {\dfrac{{\sqrt {\text{x}} }}{{\sqrt {\text{x}} {\text{ + }}\sqrt {{\text{a - x}}} }}} {\text{dx}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{a}} {\dfrac{{\sqrt {\text{x}} }}{{\sqrt {\text{x}} {\text{ + }}\sqrt {{\text{a - x}}} }}} {\text{dx}}$ ………(i)

${\text{ = }}\int_{\text{0}}^{\text{a}} {\dfrac{{\sqrt {{\text{a - x}}} }}{{\sqrt {{\text{a - x}}} {\text{ + }}\sqrt {{\text{a - (a - x)}}} }}} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{\text{a}} {\dfrac{{\sqrt {{\text{a - x}}} }}{{\sqrt {{\text{a - x}}} {\text{ + }}\sqrt {{\text{(x)}}} }}}$ …………(ii)

(i)+(ii)

\begin{align}{\text{2I = }}\int_{\text{0}}^{\text{a}} {\dfrac{{\sqrt {{\text{a - x}}} {\text{ + }}\sqrt {\text{x}} }}{{\sqrt {{\text{a - x}}} {\text{ + }}\sqrt {{\text{(x)}}} }}} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{a}} {{\text{1 }}} {\text{dx = [x]}}_{\text{0}}^{\text{a}} \hfill \\{\text{ = a}} \hfill \\{\text{I = }}\dfrac{{\text{a}}}{{\text{2}}} \hfill \\ \end{align}

18. $\mathbf{\int\limits_{\text{2}}^{\text{4}} {\left| {{\text{x - 1}}} \right|} {\text{ dx}}}$

उत्तर: ${\text{I = }}\int_{\text{2}}^{\text{4}} {\text{|}} {\text{x - 1| dx}}$

\begin{align}{\text{ - }}\int_{\text{0}}^{\text{1}} {{\text{(x - 1)}}} {\text{dx + }}\int_{\text{1}}^{\text{4}} {{\text{(x - 1)}}} {\text{dx \{ }}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x)dx = }}\int_{\text{a}}^{\text{c}} {\text{f}} {\text{(x)dx + }}\int_{\text{c}}^{\text{b}} {\text{f}} {\text{(x)dx}}\} \hfill \\{\text{ - }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ + x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - x}}} \right]_{\text{1}}^{\text{4}}\quad {\text{[(0,1) , (x - 1)}} \leqslant {\text{0 ; (1,4) , (x - 1)}} \geqslant {\text{0]}} \hfill \\\left[ {\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ - 1}}} \right){\text{ - 0}}} \right]{\text{ + }}\left( {\dfrac{{{\text{16}}}}{{\text{2}}}{\text{ - 4}}} \right){\text{ - }}\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ - 1}}} \right){\text{ = 5}} \hfill\\ \end{align}

19. दर्शाइए की $\mathbf{\int\limits_{\text{0}}^{\text{a}} {{\text{f(x)g(x)}}\;{\text{ = }}\;{\text{2}}\int\limits_{\text{0}}^{\text{a}} {{\text{f(x)dx}}} }}$ यदि ${\text{f}}$ और $\mathbf{{\text{g}}}$ को $\mathbf{{\text{f(x)}}\;{\text{ = }}\;{\text{f(a - x)}}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)g(x)dx}}$

\begin{align}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(a - x)g(a - x)dx}} \hfill \\{\text{f(a - x) = f(x) , g(x) + g(a - x) = 4 ; g(a - x) = 4 - g(x)}} \hfill \\\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(a - x)g(a - x)dx = }}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)[4 - g(x)]dx}} \hfill \\{\text{ = 4}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx - }}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)g(x)dx = 4}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx - 1}} \hfill \\{\text{2I = 4}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx}} \hfill \\{\text{I = 2}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx}} \hfill \\ \end{align}

20 एवं 21 मे सही उत्तर का चयन कीजिए:

20. $\mathbf{\int\limits_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(}}{{\text{x}}^{\text{3}}}{\text{ + xcosx + ta}}{{\text{n}}^{\text{5}}}{\text{x + 1)dx}}} }$ का मान है।

1. $\mathbf{{\text{0}}}$

2. $\mathbf{{\text{2}}}$

3. $\mathbf{{\text{1}}}$

4. $\mathbf{{{\pi }}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int\limits_{{\text{ - }}\dfrac{{\text{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(}}{{\text{x}}^{\text{3}}}{\text{ + xcosx + ta}}{{\text{n}}^{\text{5}}}{\text{x + 1)dx}}}$

\begin{align}{\text{I}}\;{\text{ = }}\;\int_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {[\left( {{{\text{x}}^{\text{3}}}{\text{ + xcosx + ta}}{{\text{n}}^{\text{5}}}{\text{x}}} \right)} {\text{dx + }}\int_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{1}} ]{\text{dx = }}{{\text{l}}_{\text{1}}}{\text{ + [x]}}_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}}{\text{ = }}{{\text{l}}_{\text{1}}}{{ + \pi }} \hfill \\{{\text{l}}_{\text{1}}}{\text{ = }}\int_{{\text{ - }}\dfrac{{{\pi }}}{{\text{2}}}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {{{\text{x}}^{\text{3}}}{\text{ + x cosx + ta}}{{\text{n}}^{\text{5}}}{\text{x}}} \right)} {\text{dx}} \hfill \\{\text{f(x) = }}\left( {{{\text{x}}^{\text{3}}}{\text{ + xcosx + ta}}{{\text{n}}^{\text{5}}}{\text{x}}} \right) \hfill \\{\text{f( - x) = }}\left( {{\text{ - }}{{\text{x}}^{\text{3}}}{\text{ - xcosx - ta}}{{\text{n}}^{\text{5}}}{\text{x}}} \right){\text{ = - f(x)}} \hfill \\ \end{align}

${\text{f(x)}}$ एक विषम फलन है

इसलिए ${{\text{I}}_1}\; = \;0$

${\text{I}}\;{\text{ = }}\;{{\text{I}}_{\text{1}}}{{ + \pi }}\;{\text{ = }}\;{t{\pi }}$

अतः विकल्प (c) सही है।

21.$\mathbf{\int\limits_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(}}\dfrac{{{\text{4 + 3sinx}}}}{{{\text{4 + 3cosx}}}}{\text{)dx}}} }$ का मान है।

1. $\mathbf{{\text{2}}}$

2. $\mathbf{\dfrac{{\text{3}}}{{\text{2}}}}$

3. $\mathbf{{\text{0}}}$

4. $\mathbf{{\text{ - 2}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int\limits_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(}}\dfrac{{{\text{4 + 3sinx}}}}{{{\text{4 + 3cosx}}}}{\text{)dx}}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4 + 3sin}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}{{{\text{4 + 3}}\left( {\dfrac{{{\pi }}}{{\text{2}}}{\text{ - x}}} \right)}}} \right)} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4 + 3cosx}}}}{{{\text{4 + 3sinx}}}}} \right)} {\text{dx}} \hfill \\{\text{ = - }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4 + 3sinx}}}}{{{\text{4 + 3cosx}}}}} \right)} \hfill \\{\text{I = - 1}} \hfill \\{\text{2I = 0}} \hfill \\{\text{I = 0}} \hfill \\ \end{align}

अतः विकल्प (c) सही है।

### प्रश्नावली A7

1. $\mathbf{\left( {{\text{1/}}\left( {{\text{x - }}{{\text{x}}^{\text{3}}}} \right)} \right)}$

उत्तर: $\left( {{\text{1/}}\left( {{\text{x - }}{{\text{x}}^{\text{3}}}} \right)} \right){\text{ = }}\left( {{\text{1/}}\left( {{\text{x}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)} \right)} \right){\text{ = (1/(x(1 - x)(1 + x)))}}$

\begin{align}{\text{(1/(x(1 - x)(1 + x))) = (A/x) + (B/(1 - x)) + (C/(1 + x))}} \hfill \\{\text{1 = A}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right){\text{ + Bx(1 + x) + Cx(1 - x)}} \hfill \\\end{align}

${\text{1}}\;{\text{ = }}\;{\text{A - A}}{{\text{x}}^{\text{2}}}{\text{ + Bx + B}}{{\text{x}}^{\text{2}}}{\text{ + Cx + C}}{{\text{x}}^{\text{2}}}$

${{\text{x}}^{\text{2}}}{\text{,}}\;{\text{x}}$ और निरंतर अवधि के गुणांक की समानता, हम प्राप्त करते है

\begin{align}{\text{ - A + B - C}}\;{\text{ = }}\;{\text{0}} \hfill \\{\text{B + C}}\;{\text{ = }}\,{\text{0}} \hfill \\{\text{A}}\;{\text{ = }}\;{\text{1}} \hfill \\ \end{align}

इन समीकरणों को हल करने पर हम प्राप्त करते है

\begin{align}{\text{A = 1,\;B = (1/2), C = ( - 1/2)}} \hfill \\{\text{(1/(x(1 - x)(1 + x))) = (1/x) + (1/(2(1 - x))) - (1/2(1 + x))}} \hfill \\\int {{\text{(1/x(}}} {\text{1 - x)(1 + x))}}\;{\text{ = }}\int {{\text{(1/x)}}} {\text{dx + (1/2)}}\int {{\text{(1/(}}} {\text{1 - x))dx - (1/2)}}\int {{\text{(1/1 + x)}}} {\text{dx}} \hfill \\{\text{ = log|x| - (1/2)log|(1 - x)| - (1/2)log|(1 + x)|}} \hfill \\{\text{ = log|x| - log}}\left| {{{{\text{(1 - x)}}}^{{\text{1/2}}}}} \right|{\text{ - log}}\left| {{{{\text{(1 + x)}}}^{{\text{1/2}}}}} \right| \hfill \\{\text{ = log}}\left| {\left( {{\text{x/}}\left( {{{{\text{(1 - x)}}}^{{\text{1/2}}}}{{{\text{(1 + x)}}}^{{\text{1/2}}}}} \right)} \right)} \right|{\text{ + C}} \hfill \\{\text{ = log}}\left| {{{\left( {{{\text{x}}^{\text{2}}}{\text{/}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)} \right)}^{{\text{1/2}}}}} \right|{\text{ + C}} \hfill \\{\text{ = (1/2)log}}\left| {{{\text{x}}^{\text{2}}}{\text{/}}\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)} \right|{\text{ + C}} \hfill \\ \end{align}

2. $\mathbf{{\text{(1/}}\sqrt {{\text{x + a}}} {\text{ + }}\sqrt {{\text{x + b}}} {\text{)}}}$

उत्तर: ${\text{(1/}}\sqrt {{\text{(x + a)}}} {\text{ + }}\sqrt {{\text{(x + b)}}} {\text{) = (1/}}\sqrt {{\text{(x + a)}}} {\text{ + }}\sqrt {{\text{(x + b)}}} {\text{) \times }}\sqrt {{\text{(x + a)}}} {\text{( - }}\sqrt {{\text{(x + b)}}} {\text{)/(}}\sqrt {{\text{(x + a)}}} {\text{ - }}\sqrt {{\text{(x + b)}}} {\text{)}}$

\begin{align}{\text{(}}\sqrt {{\text{(x + a)}}} {\text{ - }}\sqrt {{\text{(x + b)}}} {\text{)/((x + a) - (x + b))}} \hfill \\{\text{(}}\sqrt {{\text{(x + a)}}} {\text{ - }}\sqrt {{\text{(x + b)}}} {\text{)/(a - b)}} \hfill \\\int {{\text{(1/}}\sqrt {{\text{(x + a)}}} {\text{ - }}\sqrt {{\text{(x + b)}}} {\text{)dx}}} {\text{ = }}{\kern 1pt} {\text{ (1/a - b)}}\int {\text{(}} \sqrt {{\text{(x + a)}}} {\text{ - }}\sqrt {{\text{(x + b)}}} {\text{)dx}} \hfill \\{\text{ = (1/(a - b))}}\left[ {{{{\text{(x + a)}}}^{{\text{3/2}}}}{\text{/(3/2) - (x + b}}{{\text{)}}^{{\text{3/2}}}}{\text{/(3/2)}}} \right] \hfill \\{\text{ = (2/3(a - b))}}\left[ {{{{\text{(x + a)}}}^{{\text{3/2}}}}{\text{ - (x + b}}{{\text{)}}^{{\text{3/2}}}}} \right]{\text{ + C}} \hfill \\ \end{align}

3. $\mathbf{\left( {{\text{1/x}}\left( {\sqrt {\left( {{\text{ax - }}{{\text{x}}^{\text{2}}}} \right)} } \right)} \right)}$

उत्तर: माना ${\text{x = (a/t)}}$

\begin{align}{\text{dx = - }}\left( {{\text{a/}}{{\text{t}}^{\text{2}}}} \right) \hfill \\\int {\left( {{\text{1/x}}\left( {\sqrt {\left( {{\text{ax - }}{{\text{x}}^{\text{2}}}} \right)} } \right)} \right)} {\text{dx = }}\int {{\text{(1/(a/t)(}}\sqrt {{\text{(a(a/t) - (a/t}}{{\text{)}}^{\text{2}}}{\text{)( - (a/}}{{\text{t}}^{\text{2}}}{\text{)}}} } {\text{dt)}} \hfill \\{\text{ = - }}\int {{\text{(1/at)}}} {\text{ \times (1/}}\sqrt {((1/t) - (1/{t^2}))} {\text{dt}} \hfill \\{\text{ = - (1/a)}}\int {\left( {{\text{1/}}\sqrt {\left( {\left( {{{\text{t}}^{\text{2}}}{\text{/t}}} \right){\text{ - }}\left( {{{\text{t}}^{\text{2}}}{\text{/t}}} \right)} \right)} {\text{dt}}} \right.} \hfill \\{\text{ = - (1/a)(1/(}}\sqrt {{\text{(t - 1)}}} {\text{))dt}} \hfill \\ {\text{ = - (1/a)[2(}}\sqrt {\text{t}} {\text{ - 1)] + C}} \hfill \\{\text{ = - (1/a)[2(}}\sqrt {{\text{(a/x) - 1}}} {\text{)] + C}} \hfill \\ \end{align}

${\text{ = }}\;{\text{ - (2/a)[}}\sqrt {{\text{(a - x)/x}}} {\text{] + C}}$

4. $\mathbf{\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{{\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)}^{{\text{3/4}}}}} \right)} \right)}$

उत्तर: ${{\text{x}}^{{\text{ - 3}}}}$ से गुणा करना और विभाजित करना, हम प्राप्त करते है

\begin{align}\left( {{\text{x}}{{\text{ - }}^{\text{3}}}{\text{/}}\left( {{{\text{x}}^{\text{2}}}{\text{ \times x}}{{\text{ - }}^{\text{3}}}{{\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)}^{\text{3}}}{\text{/4}}} \right)} \right){\text{ = }}\left. {\left( {{\text{x}}{{\text{ - }}^{\text{3}}}\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right){\text{ - 3/4}}} \right){\text{/}}{{\text{x}}^{\text{2}}}{{\text{x}}^{\text{3}}}} \right) \hfill \\{\text{ = (}}{{\text{x}}^4}{\text{ + 1) - 3/4/(}}{{\text{x}}^5}{\text{(}}{{\text{x}}^4}{\text{) - 3/4)}} \hfill \\{\text{ = }}\left( {{\text{1/}}{{\text{x}}^{\text{5}}}} \right)\left( {\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right){\text{/}}{{\text{x}}^{\text{4}}}} \right){\text{ - 3/4}} \hfill \\{\text{ = }}\left( {{\text{1/}}{{\text{x}}^{\text{5}}}} \right)\left( {\left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{4}}}} \right)} \right){\text{ - 3/4}}} \right) \hfill \\ \end{align}

माना $\left( {{\text{1/}}{{\text{x}}^{\text{4}}}} \right){\text{ = t}}$

\begin{align}\left( {{\text{ - 4/}}{{\text{x}}^{\text{5}}}} \right){\text{dx = dt}} \hfill \\ \left( {{\text{1/}}{{\text{x}}^{\text{5}}}} \right){\text{dx = ( - dt/4)}} \hfill \\\int {\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{{\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)}^{\text{3}}}{\text{/4}}} \right)} \right)} {\text{dx = }}\int {{\text{(1/x5)}}} \left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{4}}}} \right){\text{ - 3/4}}} \right){\text{dx}} \hfill \\{\text{ = ( - 1/4)}}\int {{\text{(1 + t)}}} {\text{ - 3/4 dt}} \hfill \\{\text{ = ( - 1/4)}}\left[ {{{{\text{(1 + t)}}}^{{\text{1/4}}}}{\text{/(1/4)}}} \right]{\text{ + C}} \hfill \\ \end{align}

${\text{ = }}\;{\text{ - (1 + (1/}}{{\text{x}}^{\text{4}}}{\text{))(1/4) + C}}$

5. $\mathbf{\left( {{\text{1/}}\left( {{{\text{x}}^{{\text{1/2}}}}{\text{ + }}{{\text{x}}^{{\text{1/3}}}}} \right)} \right)}$

उत्तर: ${\text{(1/(}}{{\text{x}}^{{\text{1/2}}}}{\text{ + }}{{\text{x}}^{{\text{1/3}}}}{\text{))}}\;{\text{ = }}\;{\text{(1/(}}{{\text{x}}^{{\text{1/3}}}}{\text{(1 + }}{{\text{x}}^{{\text{1/6}}}}{\text{)}}$

माना ${\text{x}}\;{\text{ = }}\;{{\text{t}}^{\text{6}}}$

\begin{align}{\text{dx = 6}}{{\text{t}}^{\text{5}}}{\text{dt}} \hfill \\\int {\left( {{\text{1/}}\left( {{{\text{x}}^{{\text{1/2}}}}{\text{ + }}{{\text{x}}^{\text{1}}}{\text{/3}}} \right)} \right)} {\text{dx = }}\int {\left( {{\text{1/}}{{\text{x}}^{{\text{1/3}}}}\left( {{\text{1 + }}{{\text{x}}^{{\text{1/6}}}}} \right)} \right)} {\text{dx}} \hfill \\{\text{ = }}\int {\left( {{\text{6}}{{\text{t}}^{\text{5}}}{\text{dt}}} \right)} {\text{/}}\left( {{{\text{t}}^{\text{2}}}{\text{(1 + t)}}} \right){\text{dt}} \hfill \\{\text{ = 6}}\int {\left( {{{\text{t}}^{\text{3}}}{\text{/(1 + t)}}} \right)} {\text{dt}} \hfill \\\int {\left( {{\text{1/}}\left( {{{\text{x}}^{{\text{1/2}}}}{\text{ + }}{{\text{x}}^{{\text{1/3}}}}} \right)} \right)} {\text{dx = 6}}\int {\left\{ {\left( {{{\text{t}}^{\text{2}}}{\text{ - t + 1}}} \right){\text{ - (1/(1 + t))}}} \right\}} {\text{dt}} \hfill \\{\text{ = 6}}\left[ {\left( {{{\text{t}}^{\text{3}}}{\text{/3}}} \right){\text{ - }}\left( {{{\text{t}}^{\text{2}}}{\text{/2}}} \right){\text{ + t - log|1 + t|}}} \right] \hfill \\{\text{ = 2}}{{\text{x}}^{{\text{1/2}}}}{\text{ - 3}}{{\text{x}}^{1/3}}{\text{ + 6}}{{\text{x}}^{{\text{1/6}}}}{\text{ - log}}\left( {{\text{1 + }}{{\text{x}}^{{\text{1/6}}}}} \right){\text{ + C}} \hfill \\{\text{ = 2}}\sqrt {\text{x}} {\text{ - 3}}{{\text{x}}^{{\text{1/3}}}}{\text{ + 6}}{{\text{x}}^{{\text{1/6}}}}{\text{ - log}}\left( {{\text{1 + }}{{\text{x}}^{{\text{1/6}}}}} \right){\text{ + C}} \hfill \\ \end{align}

\begin{align}{\text{(5x)/}}\left( {{\text{(x + 1)}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)} \right){\text{ = (A/(x + 1)) + (Bx + C)/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right) \hfill \\{\text{5x = A}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right){\text{ + (Bx + C)(x + 1)}} \hfill \\{\text{5x = A}}{{\text{x}}^{\text{2}}}{\text{ + 9A + B}}{{\text{x}}^{\text{2}}}{\text{ + Bx + Cx + C}} \hfill \\ \end{align}

6. $\mathbf{{\text{(5x)/[(}}{{\text{x}}^2}{\text{ + 9)(x + 1)]}}}$

उत्तर: ${{\text{x}}^{\text{2}}}{\text{,}}\;{\text{x}}$ और निरंतर अवधि के गुणांक की समानता, हम प्राप्त करते है

\begin{align}{\text{A + B = 0}} \hfill \\{\text{B + C = 5}} \hfill \\{\text{9A + C = 0}} \hfill \\{\text{A = ( - 1/2), B = (1/2), C = (9/2)}} \hfill \\{\text{(5x)/}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right){\text{(x + 1)}}} \right){\text{ = ( - 1/2)(1/(x + 1)) + ((x/2) + (9/2))/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right) \hfill \\\int {{\text{(5x)}}} {\text{/}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right){\text{(x + 1)}}} \right){\text{dx = }}\int {\left\{ {{\text{( - 1/2(x + 1)) + }}\left( {{\text{(x + 9)/2}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)} \right)} \right\}} {\text{ dx}} \hfill \\{\text{ = ( - 1/2)log|x + 1| + (1/2)}}\int {{\text{x/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right){\text{dx + (9/2)}}\int {\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)} \right)} } \hfill \\ \end{align}

\begin{align}{\text{ = ( - 1/2)log|x + 1| + (1/4)}}\int {\left( {{\text{2x/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)} \right)} {\text{dx + (9/2)}}\int {\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right)} \right)} {\text{dx}} \hfill \\{\text{ = ( - 1/2)log|x + 1| + (1/4)log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right|{\text{ + (9/2) \times (1/3)ta}}{{\text{n}}^{ - 1}}{\text{(x/3)}} \hfill \\{\text{ = ( - 1/2)log|x + 1| + (1/4)log}}\left| {{{\text{x}}^{\text{2}}}{\text{ + 9}}} \right|{\text{ + (3/2)ta}}{{\text{n}}^{ - 1}}{\text{(x/3) + C}} \hfill \\ \end{align}

7. $\mathbf{{\text{(sinx/(sin(x - a)))}}}$

उत्तर: माना ${\text{x - a}}\;{\text{ = }}\;{\text{t}}$

${\text{dx}}\;{\text{ = }}\;{\text{dt}}$

\begin{align}\int {{\text{((}}} {\text{sinx)/(sin(x - a)))dx = }}\int {{\text{((sin(t + a))/ sint)}}} {\text{ dt}} \hfill \\{\text{ = }}\int {{\text{((}}} {\text{sint cosa + cost sina)/sint)dt}} \hfill \\{\text{ = }}\int {{\text{(cosa + cost sina)}}} {\text{dt}} \hfill \\{\text{ = t cosa + sina log|sint| + }}{{\text{C}}_1} \hfill \\{\text{ = (x - a)cosa + sina log|sin(x - a)| + }}{{\text{C}}_1} \hfill \\{\text{ = x cosa + sina log|sin(x - a)| - a cosa + }}{{\text{C}}_1} \hfill \\ {\text{ = sina log|sin(x - a)| + x cosa + C}} \hfill \\ \end{align}

8. $\mathbf{\left( {{{\text{e}}^{{\text{5logx}}}}{\text{ - }}{{\text{e}}^{{\text{4logx}}}}} \right){\text{/}}\left( {{{\text{e}}^{{\text{3logx}}}}{\text{ - }}{{\text{e}}^{{\text{2logx}}}}} \right)}$

उत्तर: $= \;\left( {{{\text{e}}^{{\text{4logx}}}}\left( {{{\text{e}}^{{\text{logx}}}}{\text{ - 1}}} \right)} \right){\text{/}}\left( {{{\text{e}}^{{\text{2logx}}}}\left( {{{\text{e}}^{{\text{logx}}}}{\text{ - 1}}} \right)} \right)$

\begin{align}{\text{ = }}{{\text{e}}^{{\text{2logx}}}} \hfill \\{\text{ = }}{{\text{x}}^{\text{2}}} \hfill \\\int {\left( {{{\text{e}}^{{\text{5logx}}}}{\text{ - }}{{\text{e}}^{{\text{4logx}}}}} \right)} {\text{/}}\left( {{{\text{e}}^{{\text{3logx}}}}{\text{ - }}{{\text{e}}^{{\text{2logx}}}}} \right){\text{dx = }}\int {{{\text{x}}^{\text{2}}}\;{\text{dx}}} {\text{ = (}}{{\text{x}}^{\text{3}}}{\text{/3) + C}} \hfill \\ \end{align}

9. $\mathbf{{\text{(Cosx)/(}}\sqrt {{\text{4 - si}}{{\text{n}}^2}{\text{x}}} )}$

उत्तर: माना ${\text{sinx = t}}$

\begin{align}{\text{cosx dx = dt}} \hfill \\\int {{\text{(cosx)}}} {\text{/}}\left( {\sqrt {\left( {{\text{4 - si}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)} } \right){\text{dx = }}\int {{\text{(dt)}}} {\text{/}}\left( {\sqrt {\left( {{{{\text{(2)}}}^{\text{2}}}{\text{ - (t}}{{\text{)}}^{\text{2}}}} \right)} } \right) \hfill \\{\text{ = si}}{{\text{n}}^{ - 1}}{\text{(t/2) + C}} \hfill \\{\text{ = si}}{{\text{n}}^{ - 1}}{\text{(sin(x/2)) + C}} \hfill \\ \end{align}

10. $\mathbf{\left( {{\text{si}}{{\text{n}}^{\text{8}}}{\text{x - co}}{{\text{s}}^{\text{8}}}{\text{x}}} \right){\text{/}}\left( {{\text{1 - 2si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)}$

उत्तर: ${\text{ = }}\left( {\left( {{\text{( si}}{{\text{n}}^{\text{4}}}{\text{x + co}}{{\text{s}}^{\text{4}}}{\text{x)}}\left( {{\text{si}}{{\text{n}}^{\text{4}}}{\text{x - }}} \right.} \right.} \right.\left. {\left. {\left. {{\text{co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right)} \right){\text{/}}\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} \right)$

\begin{align}{\text{ = }}\left( {\left( {\left( {{\text{si}}{{\text{n}}^{\text{4}}}{\text{x + co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right)\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x - co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} \right){\text{/}}\left( {\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x - }}} \right.} \right.} \right.\left. {\left. {\left. {{\text{si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x}}} \right){\text{ + }}\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} \right)} \right) \hfill \\{\text{ = }}\left( {\left( {{\text{si}}{{\text{n}}^{\text{4}}}{\text{x + co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right)\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x - co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} \right){\text{/}}\left( {{\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}\left( {{\text{1 - }}} \right.} \right.\left. {\left. {\left. {{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right){\text{ + co}}{{\text{s}}^{\text{2}}}{\text{x}}\left( {{\text{1 - si}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)} \right)} \right) \hfill \\{\text{ = }}\left( {\left( {{\text{ - }}\left( {{\text{si}}{{\text{n}}^{\text{4}}}{\text{x + co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right)\left( {{\text{co}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)} \right){\text{/}}\left( {{\text{si}}{{\text{n}}^{\text{4}}}{\text{x + co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right)} \right) \hfill \\{\text{ = - cos2x}}\int {\left( {\left( {{\text{si}}{{\text{n}}^{\text{8}}}{\text{x - Co}}{{\text{s}}^{\text{8}}}{\text{x}}} \right){\text{/}}\left( {{\text{1 - 2si}}{{\text{n}}^{\text{2}}}{\text{xco}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} \right)} {\text{dx}} \hfill \\ {\text{ = }}\int {{\text{( - cos2x)}}} {\text{dx = - (sin2x)/2 + C}} \hfill \\ \end{align}

11. $\mathbf{{\text{(1/(cos(x + a)cos(x + b)))}}}$

उत्तर: ${\text{sin(a - b)}}$ से गुणा और भाग करना, हम प्राप्त करते है

\begin{align}{\text{(1/(sin(a - b)))[(sin(a - b))/(cos(x + a)cos(x + b)))]}} \hfill \\{\text{ = (1/(sin(a - b)))[(sin[(x + a) - (x + b)])/(cos(x + a)cos(x + b)))]}} \hfill \\{\text{ = (1/(sin(a - b)))[((sin(x + a)cos(x + b)) - (cos(x + a)sin(x + b))/(cos(x + a)cos(x + b)))]}} \hfill \\{\text{ = (1/(sin(a - b)))[(sin(x + a))/cos(x + a)) - (sin(x + b))/cos(x + b))]}} \hfill \\{\text{ = (1/(sin(a - b)))[tan(x + a) - tan(x + b)]}} \hfill \\ \end{align}

\begin{align}\int {{\text{(1/(cos(x + a)cos(x + b)))dx}}\;{\text{ = }}\;{\text{(1/(sin(a - b)))}}\int {{\text{[tan(x + a) - tan(x + b)]}}} } \hfill \\{\text{ = }}\;{\text{(1/sin(a - b)))}}\int {{\text{[ - log}}\left| {{\text{cos(x + a)}}} \right|} {\text{ + log}}\left| {{\text{x + b}}} \right|{\text{] + C}} \hfill \\{\text{ = }}\;{\text{(1/sin(a - b)))(log}}\left| {{\text{cos(x + b)/cos(x + a)}}} \right|{\text{) + C}} \hfill \\ \end{align}

12. ${{\text{x}}^{\text{3}}}{\text{/}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{8}}}}$

उत्तर: माना ${{\text{x}}^{\text{4}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{4}}{{\text{x}}^{\text{3}}}{\text{dx = dt}} \hfill \\\int {{\text{(}}{{\text{x}}^{\text{3}}}{\text{/}}\sqrt {{\text{(1 - }}{{\text{x}}^{\text{8}}}{\text{)}}} } {\text{)dx}}\;{\text{ = }}\;{\text{(1/4)}}\int {{\text{dt/}}\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} } \hfill \\ {\text{ = (1/4)si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t) + C}} \hfill \\ {\text{ = (1/4)si}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{\text{x}}^{\text{4}}}} \right){\text{ + C}} \hfill \\ \end{align}

13. $\mathbf{{{\text{e}}^{\text{x}}}{\text{/(1 + }}{{\text{e}}^{\text{x}}}{\text{)(2 + }}{{\text{e}}^{\text{x}}}{\text{)}}}$

उत्तर: माना ${{\text{e}}^{\text{x}}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{{\text{e}}^{\text{x}}}{\text{dx}}\;{\text{ = }}\;{\text{dt}} \hfill \\\int {{{\text{e}}^{\text{x}}}{\text{/[(1 + }}{{\text{e}}^{\text{x}}}{\text{)(2 + }}{{\text{e}}^{\text{x}}}{\text{)]}}} {\text{dx}}\;{\text{ = }}\;\int {{\text{dt/[(t + 1)(t + 2)]}}} \hfill \\{\text{ = }}\;\int {{\text{[(1/(t + 1)) - (1/(t + 2))]dt}}} \hfill \\ \end{align}

\begin{align}{\text{ = log|t + 1| - log|t + 2| + C}} \hfill \\{\text{ = log|(t + 1)/(t + 2)| + C}} \hfill \\{\text{ = log}}\left| {\left( {{{\text{e}}^{\text{x}}}{\text{ + 1}}} \right){\text{/}}\left( {{{\text{e}}^{\text{x}}}{\text{ + 2}}} \right)} \right|{\text{ + C}} \hfill \\ \end{align}

14. $\mathbf{\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)} \right)}$

उत्तर: ${\text{(1/(}}{{\text{x}}^{\text{2}}}{\text{ + 1)(}}{{\text{x}}^{\text{2}}}{\text{ + 4))}}\;{\text{ = }}\;{\text{(Ax + B)/(}}{{\text{x}}^{\text{2}}}{\text{ + 1) + (Cx + D)/(}}{{\text{x}}^{\text{2}}}{\text{ + 4)}}$

\begin{align}{\text{1}}\;{\text{ = }}\;{\text{(Ax + B)(}}{{\text{x}}^{\text{2}}}{\text{ + 4) + (Cx + D)(}}{{\text{x}}^{\text{2}}}{\text{ + 1)}} \hfill \\{\text{1}}\;{\text{ = }}\;{\text{A}}{{\text{x}}^{\text{3}}}{\text{ + 4Ax + B}}{{\text{x}}^{\text{2}}}{\text{ + 4B + C}}{{\text{x}}^{\text{3}}}{\text{ + Cx + D}}{{\text{x}}^{\text{2}}}{\text{ + D}} \hfill \\ \end{align}

${{\text{x}}^3}{\text{,}}\;{{\text{x}}^{\text{2}}}{\text{,}}\;{\text{x}}$ और निरंतर अवधि के गुणांक की समानता, हम प्राप्त करते है

\begin{align}{\text{A + C = 0}} \hfill \\ {\text{B + D = 0}} \hfill \\{\text{4A + C = 0}} \hfill \\{\text{4B + D = 1}} \hfill \\{\text{A = 0, B = (1/3), C = 0, D = ( - 1/3)}} \hfill \\\left( {{\text{1/}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right)} \right){\text{ = }}\left( {{\text{1/3}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right){\text{ - }}\left( {{\text{1/3}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)} \right) \hfill \\\int {\left( {{\text{1/}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right)} \right)} {\text{dx = (1/3)}}\int {\left( {\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right)} \right){\text{dx - }}\left( {{\text{1/}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 4}}} \right)} \right){\text{dx}}} \right)} \hfill \\ \end{align}

\begin{align}{\text{ = (1/3)ta}}{{\text{n}}^{ - 1}}{\text{x - (1/3) \times (1/2)ta}}{{\text{n}}^{ - 1}}{\text{(x/2) + C}} \hfill \\{\text{ = (1/3)ta}}{{\text{n}}^{ - 1}}{\text{x - (1/6)ta}}{{\text{n}}^{ - 1}}{\text{(x/2) + C}} \hfill \\ \end{align}

15. $\mathbf{{\text{co}}{{\text{s}}^{\text{3}}}{\text{x }}{{\text{e}}^{{\text{log sinx}}}}}$

उत्तर: ${\text{ = }}\;{\text{co}}{{\text{s}}^{\text{3}}}{\text{x}}\;{\text{sinx}}$

माना ${\text{cosx}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{ - sinx dx = dt}} \hfill \\\int {{\text{co}}{{\text{s}}^{\text{3}}}} {\text{x }}{{\text{e}}^{{\text{logsinx}}}}{\text{dx = }}\int {{\text{co}}{{\text{s}}^{\text{3}}}} {\text{x sinx dx}} \hfill \\{\text{ = - }}\int {\text{t}} {\text{ dt = - }}\left( {{{\text{t}}^{\text{4}}}{\text{/4}}} \right){\text{ + C}} \hfill \\{\text{ = - }}\left( {\left( {{\text{co}}{{\text{s}}^{\text{4}}}{\text{x}}} \right){\text{/4}}} \right){\text{ + C}} \hfill \\ \end{align}

16. $\mathbf{{{\text{e}}^{{\text{3logx}}}}{\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)^{{\text{ - 1}}}}}$

उत्तर: ${\text{ = }}\left( {{{\text{x}}^{\text{3}}}} \right){\text{/}}\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)$

माना ${{\text{x}}^{\text{4}}}{\text{ + 1 = t}}$

\begin{align}{\text{4}}{{\text{x}}^{\text{3}}}{\text{ dx = dt}} \hfill \\\int {{{\text{e}}^{{\text{3logx}}}}} {\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)^{{\text{ - 1}}}}{\text{dx = }}\int {\left( {{{\text{x}}^{\text{3}}}{\text{/}}\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)} \right)} {\text{ dx}} \hfill \\ {\text{ = (1/4)}}\int {{\text{(dt/t)}}} \hfill \\\end{align}

\begin{align} {\text{ = (1/4)log|t| + C}} \hfill \\{\text{ = (1/4)log}}\left| {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right|{\text{ + C}} \hfill \\{\text{ = (1/4)log}}\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right){\text{ + C}} \hfill \\ \end{align}

17. ${{\text{f}}^{\text{1}}}{\text{(ax + b)[f(ax + b)dx}}{{\text{]}}^{\text{n}}}$

उत्तर: माना ${\text{f(ax + b) = t}}$

\begin{align}{\text{a}}{{\text{f}}^{\text{1}}}{\text{(ax + b)dx = dt}} \hfill \\\int {{{\text{f}}^{\text{1}}}} {\text{(ax + b)[f(ax + b)dx}}{{\text{]}}^{\text{n}}}{\text{dx = (1/a)}}\int {{{\text{t}}^{\text{n}}}} {\text{dt}} \hfill \\{\text{ = (1/a)}}\left[ {{{\text{t}}^{{\text{n + 1}}}}{\text{/(n + 1)}}} \right] \hfill \\{\text{ = (1/a(n + 1))(f(ax + b)}}{{\text{)}}^{{\text{n + 1}}}}{\text{ + C}} \hfill \\ \end{align}

18. $\mathbf{\left( {{\text{1/}}\left( {\sqrt {{\text{si}}{{\text{n}}^{\text{3}}}} {\text{xsin(x + a)}}} \right)} \right)}$

उत्तर: $\left( {{\text{1/}}\left( {\sqrt {{\text{si}}{{\text{n}}^{\text{3}}}} {\text{xsin(x + a)}}} \right)} \right){\text{ = }}\left( {{\text{1/}}\left( {\sqrt {{\text{si}}{{\text{n}}^{\text{3}}}} {\text{x(sinx cosa + cosx sina)}}} \right)} \right)$

\begin{align}{\text{ = }}\;{\text{1/}}\sqrt {\left( {{\text{si}}{{\text{n}}^{\text{4}}}} \right.x\;{\text{cosa + si}}{{\text{n}}^{\text{3}}}{\text{x cosx sina)}}} \hfill \\ {\text{ = cose}}{{\text{c}}^{\text{2}}}{\text{x/}}\sqrt {{\text{(cota + cotx sina)}}} \hfill \\ \end{align}

माना ${\text{cota + cotx sina = t}}$

\begin{align}{\text{ - cose}}{{\text{c}}^{\text{2}}}{\text{x sina dx = dt}} \hfill \\\int {\left( {{\text{1/}}\left( {{\text{si}}{{\text{n}}^{\text{3}}}{\text{xsin(x + a)}}} \right)} \right)} {\text{dx = }}\int {{\text{(cose}}{{\text{c}}^{\text{2}}}{\text{x/}}\sqrt {{\text{(cota + cotx}}\;{\text{sina)}}} } {\text{)dx}} \hfill \\ \end{align}

\begin{align}{\text{ = ( - 1/sin a)}}\int {{\text{(dt/}}\sqrt {\text{t}} {\text{)}}} \hfill \\{\text{ = ( - 1/sina)(2}}\sqrt {\text{t}} {\text{) + C}} \hfill \\{\text{ = ( - 1/sina)[2}}\sqrt {{\text{(cosa + cotxsina)}}} {\text{] + C}} \hfill \\{\text{ = ( - 2/sina)}}\sqrt {{\text{((sin(x + a))/sinx)}}} {\text{ + C}} \hfill \\ \end{align}

19. $\mathbf{\dfrac{{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ - co}}{{\text{s}}^{{\text{ - 1}}}}\sqrt {\text{x}} }}{{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\sqrt {\text{x}} }}{\text{, x}} \in {\text{[0,1]}}}$

उत्तर: ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ + co}}{{\text{s}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ = }}\dfrac{{\text{\pi }}}{{\text{2}}}$  ………..(i)

${\text{co}}{{\text{s}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ = }}\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}}$ …………….(ii)

\begin{align}\therefore \;\int {\dfrac{{{\text{si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ - }}\left( {\dfrac{{\text{\pi }}}{{\text{2}}}{\text{ - si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} } \right)}}{{{\text{\pi /2}}}}} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{\pi }}}\int {\left( {{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{ - }}\dfrac{{\text{\pi }}}{{\text{2}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{\text{4}}}{{{\pi }}}\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} \sqrt {\text{x}} {\text{dx - }}\int {\text{1}} {\text{ dx}} \hfill \\ \end{align}

माना ${\text{x = si}}{{\text{n}}^{\text{2}}}{\text{\theta }}$

${\text{dx = 2 sin\theta cos\theta d\theta }}$

\begin{align}= \;\dfrac{{\text{4}}}{{{\pi }}}\int {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{(sin\theta )2 sin\theta cos\theta d\theta - x + C}} \hfill \\{\text{ = }}\dfrac{{\text{4}}}{{{\pi }}}\int {{\text{\theta }}} {\text{sin2\theta d\theta - x + C}} \hfill \\{\text{ = }}\dfrac{{\text{4}}}{{{\pi }}}\left[ {\dfrac{{{\text{ - \theta cos2\theta }}}}{{\text{2}}}{\text{ + }}\int {\text{1}} {\text{ \times }}\dfrac{{{\text{cos2\theta }}}}{{\text{2}}}{\text{\;d\theta }}} \right]{\text{ - x + C}} \hfill \\ \end{align}

\begin{align}= \;\dfrac{{\text{4}}}{{{\pi }}}\left[ {\dfrac{{{\text{ - \theta cos2\theta }}}}{{\text{2}}}{\text{ + }}\dfrac{{{\text{sin2\theta }}}}{{\text{4}}}} \right]{\text{ - x + C}} \hfill \\{\text{ = }}\dfrac{{\text{4}}}{{{{4 \times \pi }}}}\left[ {{\text{ - 2si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} {\text{(1 - 2x) + 2}}\sqrt {\text{x}} \sqrt {{\text{1 - x}}} } \right]{\text{ - x + C}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{{\pi }}}\left[ {\sqrt {{\text{x - }}{{\text{x}}^{\text{2}}}} {\text{ - (1 - 2x)si}}{{\text{n}}^{{\text{ - 1}}}}\sqrt {\text{x}} } \right]{\text{ - x + C}} \hfill \\ \end{align}

20. $\sqrt {{\text{(1 - }}\sqrt {\text{x}} {\text{)/(1 + }}\sqrt {\text{x}} {\text{)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;\int {\sqrt {{\text{(1 - }}\sqrt {\text{x}} {\text{)/(1 + }}\sqrt {\text{x}} {\text{)}}} } {\text{dx}}$

माना ${\text{x}}\;{\text{ = }}\;{\text{co}}{{\text{s}}^{\text{2}}}{\text{\theta }}$

\begin{align}{\text{dx = - 2 sin\theta cos\theta d\theta }} \hfill \\{\text{I = }}\int {\sqrt {{\text{(1 - cos\theta )/(1 + cos\theta ) \times ( - 2sin\theta cos\theta )}}} {\text{d\theta }}} \hfill \\{\text{ = - }}\int {{\text{tan}}} {\text{(\theta /2) \times 2 sin\theta cos\theta d\theta }} \hfill \\ \end{align}

\begin{align}{\text{ = - 2}}\int {{\text{(sin(}}} {\text{\theta /2)/cos(\theta /2))(2sin(\theta /2)cos(\theta /2))cos\theta d\theta }} \hfill \\{\text{I = - 4}}\int {\left( {{\text{2si}}{{\text{n}}^{\text{2}}}{\text{(\theta /2)co}}{{\text{s}}^{\text{2}}}{\text{(\theta /2) - si}}{{\text{n}}^{\text{2}}}{\text{(\theta /2)}}} \right)} {\text{d\theta }} \hfill \\ \end{align}

${\text{ = - 8}}\int {{\text{si}}{{\text{n}}^{\text{2}}}} {\text{(\theta /2)co}}{{\text{s}}^{\text{2}}}{\text{(\theta /2)d\theta + 4}}\int {{\text{si}}{{\text{n}}^{\text{2}}}} {\text{(\theta /2)d\theta }}$

\begin{align}{\text{ = - 2}}\int {{\text{((}}} {\text{1 - cos2\theta )/2)d\theta + 4}}\int {{\text{((}}} {\text{1 - cos\theta )/2)d\theta + C}} \hfill \\{\text{ = \theta + (sin2\theta /2) + 2\theta - 2sin\theta + C}} \hfill \\{\text{ = \theta + }}\left( {\sqrt {\left( {{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{\theta }}} \right)} } \right){\text{ \times cos\theta - 2}}\left( {\sqrt {\left( {{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{\theta }}} \right)} } \right){\text{ + C}} \hfill \\{\text{ = - 2(}}\sqrt {{\text{(1 - x)}}} {\text{ + co}}{{\text{s}}^{ - 1}}\sqrt {\text{x}} {\text{ + }}\sqrt {\left( {{\text{x - }}{{\text{x}}^{\text{2}}}} \right)} {\text{ + C}} \hfill \\ \end{align}

21. ${\text{I = }}\int {{\text{(2 + sin2x)}}} {\text{/(1 + cos2x)}}{{\text{e}}^{\text{x}}}$

उत्तर: $= \;\int {\left( {{\text{(2 + 2sinxcosx)/2co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} {\text{ }}{{\text{e}}^{\text{x}}}$

$= \;\int {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x + tanx}}} \right)} {{\text{e}}^{\text{x}}}$

माना ${\text{f(x) = tanx , }}{{\text{f}}^'}{\text{(x) = se}}{{\text{c}}^{\text{2}}}{\text{x}}$

\begin{align}{\text{I}}\;{\text{ = }}\;\int {{\text{[f(x) + }}{{\text{f}}^{\text{'}}}{\text{(x)}}} {\text{]}}{{\text{e}}^{\text{x}}}{\text{dx}} \hfill \\{\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{f(x) + C}} \hfill \\{\text{ = }}\;{{\text{e}}^{\text{x}}}{\text{tanx + C}} \hfill \\ \end{align}

22. $\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right){\text{/(x + 1}}{{\text{)}}^{\text{2}}}{\text{(x + 2)}}} \right)$

उत्तर: $\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right){\text{/(x + 1}}{{\text{)}}^{\text{2}}}{\text{(x + 2)}}} \right){\text{ = (A/(x + 1)) + }}\left( {{\text{B/(x + 1}}{{\text{)}}^{\text{2}}}} \right){\text{ + (C/(x + 2))}}$

\begin{align}{{\text{x}}^{\text{2}}}{\text{ + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C}}\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right) \hfill \\{{\text{x}}^{\text{2}}}{\text{ + x + 1 = (A + C)}}\left( {{{\text{x}}^{\text{2}}}} \right){\text{ + (3A + B + 2C)x + (2A + 2B + C)}} \hfill \\ \end{align}

${{\text{x}}^{\text{2}}}{\text{,}}\;{\text{x}}$ और निरंतर अवधि के गुणांक की समानता, हम प्राप्त करते है

\begin{align}{\text{A + C = 1}} \hfill \\ {\text{3A + B + 2C = 1}} \hfill \\ \end{align}

\begin{align}{\text{2A + 2B + C = 1}} \hfill \\\therefore \;{\text{A = - 2, B = 1, C = 3}} \hfill \\\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right){\text{/}}\left( {{{{\text{(x + 1)}}}^{\text{2}}}{\text{(x + 2)}}} \right)} \right){\text{ = ( - 2/(x + 1)) + (3/(x + 2)) + }}\left( {{\text{1/(x + 1}}{{\text{)}}^{\text{2}}}} \right) \hfill \\\int {\left( {{{\text{x}}^{\text{2}}}{\text{ + x + 1}}} \right)} {\text{/}}\left( {{{{\text{(x + 1)}}}^{\text{2}}}{\text{(x + 2)}}} \right){\text{dx = - 2}}\int {{\text{(1/(}}} {\text{x + 1))dx + 3}}\int {{\text{(1/(}}} {\text{x + 2))dx + }}\int {\left( {{\text{1/(x + 1}}{{\text{)}}^{\text{2}}}} \right)} {\text{dx}} \hfill \\{\text{ = - 2log|x + 1| + 3|og|x + 2}}\mid {\text{ - (1/(x + 1)) + C}} \hfill \\ \end{align}

23. ${\text{ta}}{{\text{n}}^{ - 1}}\sqrt {{\text{(1 - x)/(1 + x)}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ ${\text{ta}}{{\text{n}}^{ - 1}}\sqrt {{\text{(1 - x)/(1 + x)}}}$

माना ${\text{x}}\;{\text{ = }}\;{\text{cos\theta }}$

\begin{align}{{dx = - sin\theta d\theta }} \hfill \\{\text{I}}\;{\text{ = }}\;\int {{{\tan }^{ - 1}}\sqrt {{\text{(1 - cos\theta )/(1 + cos\theta )}}} } {{ \times ( - sin\theta d\theta )}} \hfill \\{\text{ = - }}\int {{\text{(ta}}{{\text{n}}^{ - 1}}{\text{tan(}}} {{\theta /2)sin\theta )d\theta }} \hfill \\{\text{ = ( - 1/2)}}\left[ {{{\theta \times ( - cos\theta ) - }}\int {{\text{(1}}{\text{.(}}} {{ - cos\theta ))d\theta }}} \right] \hfill \\{{ = (1/2)\theta cos\theta - (1/2)sin\theta }} \hfill \\ \end{align}

${\text{ = }}\;{\text{(1/2)(xco}}{{\text{s}}^{{\text{ - 1}}}}{\text{x - (}}\sqrt {{\text{1 - }}{{\text{x}}^{\text{2}}}} {\text{)) + C}}$

24. $\sqrt {\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)} \left[ {{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - 2logx}}} \right]{\text{/x}}$

उत्तर: $\sqrt {\left( {{{\text{x}}^{\text{4}}}{\text{ + 1}}} \right)} \left[ {{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - 2logx}}} \right]{\text{/x = }}\sqrt {\left. {{\text{(}}{{\text{x}}^{\text{2}}}{\text{ + 1}}){\text{/}}{{\text{x}}^{\text{4}}}} \right)} \left[ {{\text{log}}\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{ - log}}{{\text{x}}^{\text{2}}}} \right]{\text{ }}$

$= \;[(\sqrt {{x^2} + 1} )/{x^4}]\left[ {{\text{log}}\left( {\left( {{{\text{x}}^{\text{2}}}{\text{ + 1}}} \right){\text{/}}{{\text{x}}^{\text{2}}}} \right)} \right]$

\begin{align}{\text{ = }}\left( {{\text{1/}}{{\text{x}}^{\text{3}}}} \right)(\sqrt {{\text{(}}{{\text{x}}^2}{\text{ + 1)/}}{{\text{x}}^2}} {\text{)log}}\left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{2}}}} \right)} \right) \hfill \\{\text{ = }}\left( {{\text{1/}}{{\text{x}}^{\text{3}}}} \right){\text{(}}\sqrt {(1 + (1/{x^2}))} )\log (1 + (1/{x^2})) \hfill \\ \end{align}

माना ${\text{(1 + (1/}}{{\text{x}}^{\text{2}}}{\text{))}}\;{\text{ = }}\;{\text{t}}$

\begin{align}\left( {{\text{ - 2/}}{{\text{x}}^{\text{3}}}} \right){\text{dx = dt}} \hfill \\{\text{I = }}\int {\left( {{\text{1/}}{{\text{x}}^{\text{3}}}} \right)} \sqrt {\left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{2}}}} \right)} \right)} {\text{log}}\left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{2}}}} \right)} \right){\text{dx}} \hfill \\{\text{ = ( - 1/2)}}\int {\sqrt {\text{t}} } {\text{ logt dt}} \hfill \\{\text{I = ( - 1/2)[logt)}}\sqrt {\text{t}} {\text{dt - \{ (dt/t)logt}}\sqrt {\text{t}} {\text{dt\} dt]}} \hfill \\{\text{ = ( - 1/2)}}\left[ {{\text{logt \times (}}\sqrt {\text{t}} \sqrt {\text{t}} \sqrt {\text{t}} {\text{)/(3/2) - }}\int {{\text{(1/t)}}} {\text{(}}\sqrt {\text{t}} \sqrt {\text{t}} \sqrt {\text{t}} {\text{)/(3/2)dt}}} \right. \hfill \\{\text{ = ( - 1/2)}}\left[ {{\text{(2/3)}}{{\text{t}}^{\text{3}}}{\text{/(2logt) - (4/9)}}{{\text{t}}^{\text{3}}}{\text{/2}}} \right] \hfill \\{\text{ = ( - 1/3)}}\left( {{\text{1 + }}{{\left( {{\text{1/}}{{\text{x}}^{\text{2}}}} \right)}^{\text{3}}}{\text{/2}}} \right)\left[ {{\text{log}}\left( {{\text{1 + }}\left( {{\text{1/}}{{\text{x}}^{\text{2}}}} \right)} \right){\text{ - (2/3)}}} \right]{\text{ + C}} \hfill \\ \end{align}

### 25 से 33 तक के प्रश्नों मे निश्चित समाकलनों का मान ज्ञात कीजिए

25. $\int_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }} {{{\text{e}}^{\text{x}}}} \left( {\dfrac{{{\text{1 - sinx}}}}{{{\text{1 - cosx}}}}} \right){\text{dx}}$

उत्तर: $= \;\int_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }} {{{\text{e}}^{\text{x}}}} \left( {\dfrac{{{\text{1 - 2sin}}\dfrac{{\text{x}}}{{\text{2}}}{\text{2cos}}\dfrac{{\text{x}}}{{\text{2}}}}}{{{\text{si}}{{\text{n}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}} \right){\text{dx}}$

${\text{ = }}\int_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }} {{{\text{e}}^{\text{x}}}} \left( {\dfrac{{{\text{cose}}{{\text{c}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}}}{{\text{2}}}} \right){\text{dx}}$

माना ${\text{f(x) = - cot}}\dfrac{{\text{x}}}{{\text{2}}}$

\begin{align}{{\text{f}}^'}{\text{(x) = - }}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cose}}{{\text{c}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}}} \right){\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{cose}}{{\text{c}}^{\text{2}}}\dfrac{{\text{x}}}{{\text{2}}} \hfill \\{\text{I = }}\int_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }} {\left( {{\text{f(x) + }}{{\text{f}}^'}{\text{(x)}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\left[ {{{\text{e}}^{\text{x}}}{\text{f(x)}}} \right]_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }}{\text{ = - }}\left[ {{{\text{e}}^{\text{x}}}{\text{cot}}\dfrac{{\text{x}}}{{\text{2}}}} \right]_{\dfrac{{{\pi }}}{{\text{2}}}}^{{\pi }} \hfill \\{\text{ = }}\left[ {{{\text{e}}^{{\pi }}}{\text{cot}}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - }}{{\text{e}}^{\dfrac{{{\pi }}}{{\text{2}}}}}{\text{cot}}\dfrac{{{\pi }}}{{\text{4}}}} \right] \hfill \\ {\text{ = }}{{\text{e}}^{\dfrac{{{\pi }}}{{\text{2}}}}} \hfill \\ \end{align}

26. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {\dfrac{{{\text{sinx cosx}}}}{{{\text{co}}{{\text{s}}^{\text{4}}}{\text{x + si}}{{\text{n}}^{\text{4}}}{\text{x}}}}} \right)} {\text{dx}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {\dfrac{{{\text{sinx cosx}}}}{{{\text{co}}{{\text{s}}^{\text{4}}}{\text{x + si}}{{\text{n}}^{\text{4}}}{\text{x}}}}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\int\limits_0^{\dfrac{{{\pi }}}{{\text{4}}}} {\dfrac{{\dfrac{{{\text{sinx}}\;{\text{cosx}}}}{{{\text{co}}{{\text{s}}^{\text{4}}}{\text{x}}}}}}{{\dfrac{{{\text{co}}{{\text{s}}^{\text{4}}}{\text{x + si}}{{\text{n}}^{\text{4}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{4}}}{\text{x}}}}}}\;} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {\dfrac{{{\text{tanx se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{1 + ta}}{{\text{n}}^{\text{4}}}{\text{x}}}}} \right)} {\text{dx}} \hfill \\ \end{align}

माना ${\text{ta}}{{\text{n}}^{\text{2}}}{\text{x = t}}$

\begin{align}{\text{2 tanx se}}{{\text{c}}^{\text{2}}}{\text{x dx = dt}} \hfill \\{\text{x = 0, t = 0}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{4}}}{\text{, t = 1}} \hfill \\{\text{I = }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} \right)} {\text{dt}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]_{\text{0}}^{\text{1}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{1 - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{0}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\left[ {\dfrac{{{\pi }}}{{\text{4}}}} \right]{\text{ = }}\dfrac{{{\pi }}}{{\text{8}}} \hfill \\ \end{align}

27. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x + 4si}}{{\text{n}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x + 4si}}{{\text{n}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x + 4}}\left( {{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4 - 3co}}{{\text{s}}^{\text{2}}}{\text{x - 4}}}}{{{\text{4 - 3co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4 - 3co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{{\text{4 - 3co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx + }}\dfrac{{\text{1}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{\text{4}}}{{{\text{4 - 3co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{(1)}}} {\text{dx + }}\dfrac{{\text{1}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{4se}}{{\text{c}}^{\text{2}}}{\text{x - 3}}}}} \right)} {\text{dx}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[x]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{4se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{{\text{4}}\left( {{\text{1 - ta}}{{\text{n}}^{\text{2}}}{\text{x}}} \right){\text{ - 3}}}}} \right)} {\text{dx}} \hfill \\{\text{I = - }}\dfrac{{{\pi }}}{{\text{6}}}{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{\left( {{\text{1 + 4ta}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)}}} \right)} {\text{dx}} \hfill \\ \end{align}

${{\text{I}}_1}{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{\left( {{\text{1 + 4ta}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)}}} \right)} {\text{dx}}$

माना ${\text{tanx = t}}$

\begin{align}{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x dx = dt}} \hfill \\{\text{x = 0, t = 0}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{,t = }}\infty \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{{\text{2se}}{{\text{c}}^{\text{2}}}{\text{x}}}}{{\left( {{\text{1 + 4ta}}{{\text{n}}^{\text{2}}}{\text{x}}} \right)}}} \right)} {\text{dx = }}\dfrac{{\text{2}}}{{\text{3}}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} \right)} {\text{dt}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]_{\text{0}}^\infty \hfill \\{\text{ = }}\;\dfrac{{\text{2}}}{{\text{3}}}{\text{ \times }}\dfrac{{{\pi }}}{{\text{2}}}\;{\text{ = }}\;\dfrac{{{\pi }}}{{\text{3}}} \hfill \\{\text{I}}\;{\text{ = }}\;{\text{ - }}\dfrac{{{\pi }}}{{\text{6}}}{\text{ + }}\dfrac{{\text{\pi }}}{{\text{3}}}\;{\text{ = }}\;\dfrac{{{\pi }}}{{\text{6}}} \hfill \\ \end{align}

28. $\int_{\dfrac{{{\pi }}}{{\text{6}}}}^{\dfrac{{{\pi }}}{{\text{3}}}} {\left( {\dfrac{{{\text{sinx + cosx}}}}{{\sqrt {{\text{sin2x}}} }}} \right)} {\text{dx}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\dfrac{{{\pi }}}{{\text{6}}}}^{\dfrac{{{\pi }}}{{\text{3}}}} {\left( {\dfrac{{{\text{sinx + cosx}}}}{{\sqrt {{\text{sin2x}}} }}} \right)} {\text{dx}}$

${\text{ = }}\int_{\dfrac{{{\pi }}}{{\text{6}}}}^{\dfrac{{{\pi }}}{{\text{3}}}} {\left( {\dfrac{{{\text{sinx + cosx}}}}{{\sqrt {{\text{1 - (sinx - cosx}}{{\text{)}}^{\text{2}}}} }}} \right)} {\text{dx}}$

माना ${\text{sinx - cosx = t}}$

\begin{align}{\text{(sinx + cosx)dx = dt}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{6}}}{\text{, t = }}\dfrac{{{\text{1 - }}\sqrt {\text{3}} }}{{\text{2}}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{3}}}{\text{, t = }}\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}} \hfill \\ \end{align}

${\text{I = }}\int_{\dfrac{{{\text{1 - }}\sqrt {\text{3}} }}{{\text{2}}}}^{\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}}} {\left( {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} }}} \right)} {\text{dt}}$

${\text{I = }}\int_{{\text{ - }}\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}}}^{\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}}} {\left( {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} }}} \right)} {\text{dt}}$

$\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} }}$ यह एक सम फलन है

हमलोग जानते है कि

\begin{align}\int_{{\text{ - a}}}^{\text{a}} {\text{f}} {\text{(x)dx = 2}}\int_{\text{0}}^{\text{a}} {\text{f}} {\text{(x)dx}} \hfill \\{\text{I = 2}}\int_{\text{0}}^{\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}}} {\left( {\dfrac{{\text{1}}}{{\sqrt {{\text{1 - }}{{\text{t}}^{\text{2}}}} }}} \right)} {\text{dt}} \hfill \\{\text{I = }}\left[ {{\text{2si}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]_{\text{0}}^{\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}}} \hfill \\{\text{I = 2si}}{{\text{n}}^{{\text{ - 1}}}}\dfrac{{\sqrt {\text{3}} {\text{ - 1}}}}{{\text{2}}} \hfill \\ \end{align}

29. $\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{\text{1}}}{{\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {\text{x}} }}} \right)} {\text{dx}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{\text{1}}}{{\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {\text{x}} }}} \right)} {\text{dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{\sqrt {{\text{1 + x}}} {\text{ + }}\sqrt {\text{x}} }}{{{\text{(}}\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {\text{x}} {\text{)(}}\sqrt {{\text{1 + x}}} {\text{ - }}\sqrt {\text{x}} {\text{)}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{1}} {\left( {\dfrac{{\sqrt {{\text{1 + x}}} {\text{ + }}\sqrt {\text{x}} }}{{{\text{(1 + x - x)}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\text{1}} {{\text{(}}\sqrt {{\text{1 + x}}} {\text{)}}} {\text{dx + }}\int_{\text{0}}^{\text{1}} {{\text{(}}\sqrt {\text{x}} {\text{)}}} {\text{dx}} \hfill \\\end{align}

\begin{align}{\text{ = }}\left[ {\dfrac{{\text{2}}}{{\text{3}}}{{{\text{(1 + x)}}}^{\dfrac{{\text{2}}}{{\text{3}}}}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\left[ {\dfrac{{\text{2}}}{{\text{3}}}{{{\text{(x)}}}^{\dfrac{{\text{2}}}{{\text{3}}}}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}\left[ {{{{\text{(2)}}}^{\dfrac{{\text{2}}}{{\text{3}}}}}{\text{ - 1}}} \right]{\text{ + }}\dfrac{{\text{2}}}{{\text{3}}}{\text{[1]}} \hfill \\{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}{{\text{(2)}}^{\dfrac{{\text{2}}}{{\text{3}}}}} \hfill \\{\text{I = }}\dfrac{{{\text{4}}\sqrt {\text{2}} }}{{\text{3}}} \hfill \\ \end{align}

30. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {\dfrac{{{\text{sinx + cosx}}}}{{{\text{9 + 16sin2x}}}}} \right)} {\text{dx}}$

उत्तर: माना ${\text{sinx - cosx}}\;{\text{ = }}\;{\text{t}}$

\begin{align}{\text{(sinx + cosx)dx = dt}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{4}}}{\text{, t = 0}} \hfill \\{\text{x = 0, t = - 1}} \hfill \\{{\text{(sinx - cosx)}}^{\text{2}}}{\text{ = (t}}{{\text{)}}^{\text{2}}} \hfill \\{\text{si}}{{\text{n}}^{\text{2}}}{\text{x + co}}{{\text{s}}^{\text{2}}}{\text{x - 2sinx cosx = (t}}{{\text{)}}^{\text{2}}} \hfill \\{\text{sin2x = 1 - (t}}{{\text{)}}^{\text{2}}} \hfill \\{\text{I = }}\int_{{\text{ - 1}}}^{\text{0}} {\left( {\dfrac{{\text{1}}}{{{\text{9 + 16}}\left( {{\text{1 - }}{{\text{t}}^{\text{2}}}} \right)}}} \right)} {\text{dt = }}\int_{{\text{ - 1}}}^{\text{0}} {\left( {\dfrac{{\text{1}}}{{\left. {{\text{25 - 16}}{{\text{t}}^{\text{2}}}} \right)}}} \right)} {\text{dt}} \hfill \\{\text{ = }}\int_{{\text{ - 1}}}^{\text{0}} {\left( {\dfrac{{\text{1}}}{{\left. {{{{\text{(5)}}}^{\text{2}}}{\text{ - (4t}}{{\text{)}}^{\text{2}}}} \right)}}} \right)} {\text{dt}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{\text{1}}}{{\text{4}}}\left[ {\dfrac{{\text{1}}}{{{\text{2(5)}}}}{\text{log}}\dfrac{{{\text{|5 + 4t|}}}}{{{\text{|5 - 4t|}}}}} \right]_{{\text{ - 1}}}^{\text{0}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{40}}}}\left[ {{\text{log1 - log}}\dfrac{{\text{1}}}{{\text{9}}}} \right] \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{{\text{40}}}}{\text{log9}} \hfill \\ \end{align}

31. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{sin}}} {\text{2x ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(sinx)dx}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{sin}}} {\text{2x ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(sinx)dx = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\text{2}} {\text{ sinx cosx ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(sinx)dx}}$

माना ${\text{sinx = t}}$

\begin{align}{\text{(cosx)dx = dt}} \hfill \\ {\text{x = 0, t = 0}} \hfill \\{\text{x = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{, t = 1}} \hfill \\{\text{I = }}\int_{\text{0}}^{\text{1}} {{\text{t }}} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)dt}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int {{\text{t }}} {\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)dt}} \hfill \\{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\int {\text{t}} {\text{ dt - }}\int {\left\{ {\dfrac{{\text{d}}}{{{\text{dt}}}}\left( {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right)\int {{\text{t }}} {\text{dt}}} \right\}} {\text{dt}} \hfill \\{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\int {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} \dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{dt}} \hfill \\{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{{\text{1 + }}{{\text{t}}^{\text{2}}}{\text{ - 1}}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} {\text{dt}} \hfill \\ \end{align}

\begin{align}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}\int {\text{1}} {\text{ dt + }}\dfrac{{\text{1}}}{{\text{2}}}\int {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} {\text{ dt}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{t}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}} \hfill \\{\text{I = }}\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(t)}}\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}{\text{ - }}\dfrac{{\text{t}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t}}} \right]_{\text{0}}^{\text{1}} \hfill \\{\text{ = }}\left[ {\dfrac{{{\pi }}}{{\text{4}}}{\text{ - 1 + }}\dfrac{{{\pi }}}{{\text{4}}}} \right] \hfill \\{\text{ = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 1}} \hfill \\ \end{align}

32. $\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{\text{x tanx}}}}{{{\text{secx + tanx}}}}} \right)} {\text{dx}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{\text{x tanx}}}}{{{\text{secx + tanx}}}}} \right)} {\text{dx}}$ ………………(i)

${\text{I = }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{{(\pi - x)tan(\pi - x)}}}}{{{\text{sec(\pi - x) + tan(\pi - x)}}}}} \right)} {\text{dx = }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{{ - (\pi - x)tan(x)}}}}{{{\text{ - sec(x) + tan(x)}}}}} \right)} {\text{dx}}$

${\text{ = }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{{(\pi - x)tan(x)}}}}{{{\text{sec(x) + tan(x)}}}}} \right)} {\text{dx}}$ …………………(ii)

(i)+(ii)

\begin{align}{\text{2I = }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{{\pi tanx}}}}{{{\text{sec(x) + tan(x)}}}}} \right)} {\text{dx}} \hfill \\{\text{ = \pi }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{\dfrac{{{\text{sinx}}}}{{{\text{cosx}}}}}}{{\dfrac{{\text{1}}}{{{\text{cosx}}}}{\text{ + }}\dfrac{{{\text{sinx}}}}{{{\text{cosx}}}}}}} \right)} {\text{dx}} \hfill \\ \end{align}

\begin{align}{{ = \pi }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{{\text{sinx + 1 - 1}}}}{{{\text{1 + sinx}}}}} \right)} {\text{dx}} \hfill \\{{ = \pi }}\int_{\text{0}}^{{\pi }} {{\text{(1)}}} {\text{dx - \pi }}\int_{\text{0}}^{{\pi }} {\left( {\dfrac{{\text{1}}}{{{\text{1 + sinx}}}}} \right)} {\text{dx}} \hfill \\{{ = \pi [x]}}_{\text{0}}^{{\pi }}{{ - \pi }}\int_{\text{0}}^{\text{\pi }} {\left( {\dfrac{{{\text{1 - sinx}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}} \right)} {\text{dx}} \hfill \\{\text{ = }}{{{\pi }}^{\text{2}}}{\text{ - \pi }}\int_{\text{0}}^{{\pi }} {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - tanx secx}}} \right)} {\text{dx}} \hfill \\{\text{ = }}{{{\pi }}^{\text{2}}}{{ - \pi [tanx - secx]}}_{\text{0}}^{{\pi }}{\text{ = }}{{{\pi }}^{\text{2}}}{{ - 2\pi }} \hfill \\{\text{I = }}\dfrac{{{\pi }}}{{\text{2}}}{{(\pi - 2)}} \hfill \\ \end{align}

33. $\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 1| + |x - 2| + |x - 3|dx}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 1| + |x - 2| + |x - 3|dx}}$

\begin{align}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 1| dx + }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 2| dx + }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 3| dx}} \hfill \\{\text{I = }}{{\text{l}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 1| dx}} \hfill \\{\text{(x - 1)}} \geqslant {\text{0, 1}} \leqslant {\text{x}} \leqslant {\text{4}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 1| dx}} \hfill \\ \end{align}

\begin{align}{\text{ = }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - x}}} \right]_{\text{1}}^{\text{4}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\dfrac{{\text{9}}}{{\text{2}}} \hfill \\{{\text{I}}_{\text{1}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 2| dx}} \hfill \\{\text{(x - 2)}} \geqslant {\text{0, 2}} \leqslant {\text{x}} \leqslant {\text{4 ; (x - 2)}} \leqslant {\text{0, 1}} \leqslant {\text{x}} \leqslant {\text{2}} \hfill \\{{\text{I}}_{\text{2}}}{\text{ = }}\int_{\text{1}}^{\text{2}} {{\text{(2 - x)}}} {\text{ dx + }}\int_{\text{2}}^{\text{4}} {{\text{(x - 2)}}} {\text{ dx}} \hfill \\{\text{ = }}\left[ {{\text{2x - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right]_{\text{1}}^{\text{2}}{\text{ + }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 2x}}} \right]_{\text{2}}^{\text{4}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 2 = }}\dfrac{{\text{5}}}{{\text{2}}} \hfill \\{{\text{l}}_{\text{3}}}{\text{ = }}\int_{\text{1}}^{\text{4}} {\text{|}} {\text{x - 3| dx}} \hfill \\{\text{(x - 3)}} \geqslant {\text{0, 3}} \leqslant {\text{x}} \leqslant {\text{4 ; (x - 3)}} \leqslant {\text{0, 1}} \leqslant {\text{x}} \leqslant {\text{3}} \hfill \\ \int_{\text{1}}^{\text{3}} {{\text{(3 - x)}}} {\text{dx + }}\int_{\text{3}}^{\text{4}} {{\text{(x - 3)}}} {\text{dx = }}\left[ {{\text{3x - }}\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}} \right]_{\text{1}}^{\text{3}}{\text{ + }}\left[ {\dfrac{{{{\text{x}}^{\text{2}}}}}{{\text{2}}}{\text{ - 3x}}} \right]_{\text{3}}^{\text{4}} \hfill \\{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ + 2 = }}\dfrac{{\text{5}}}{{\text{2}}} \hfill \\{\text{I = }}{{\text{l}}_{\text{1}}}{\text{ + }}{{\text{I}}_{\text{2}}}{\text{ + }}{{\text{I}}_{\text{3}}} \hfill \\{\text{I = }}\dfrac{{\text{9}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{ + }}\dfrac{{\text{5}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{19}}}}{{\text{2}}} \hfill \\ \end{align}

### निम्नलिखित को सिद्ध कीजिए ( 34 से 39 तक )

34. $\int_{\text{1}}^{\text{3}} {\left( {\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{(x + 1)}}}}} \right)} {\text{dx = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ + log}}\dfrac{{\text{2}}}{{\text{3}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{1}}^{\text{3}} {\left( {\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{(x + 1)}}}}} \right)} {\text{dx}}$

\begin{align}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{(x + 1)}}}}{\text{ = }}\dfrac{{\text{A}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{B}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{c}}}{{{\text{x + 1}}}}{\text{ = }}\dfrac{{{\text{Ax(x + 1) + B(x + 1)c}}{{\text{x}}^{\text{2}}}}}{{{{\text{x}}^{\text{2}}}{\text{(x + 1)}}}} \hfill \\ {\text{1 = Ax(x + 1) + B(x + 1) + c}}\left( {{{\text{x}}^{\text{2}}}} \right) \hfill \\{\text{1 = A}}{{\text{x}}^{\text{2}}}{\text{ + Ax + bx + Cx}} \hfill \\ \end{align}

दोनों और ${\text{x}}$ और अचल पद की तुलना करने पर हमे मिलता है

${\text{A + C = 0, \;A + B = 0, B = 1}}$

इन सभी पदों का उकेल करने पर हमे मिलता है

\begin{align}{\text{A = - 1, B = 1, C = 1}} \hfill \\\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}{\text{(x + 1)}}}}{\text{ = }}\dfrac{{{\text{ - 1}}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{x + 1}}}} \hfill \\{\text{I = }}\int_{\text{1}}^{\text{3}} {\left\{ {\dfrac{{{\text{ - 1}}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{{{\text{x}}^{\text{2}}}}}{\text{ + }}\dfrac{{\text{1}}}{{{\text{(x + 1)}}}}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left[ {{\text{ - logx - }}\dfrac{{\text{1}}}{{\text{x}}}{\text{ + log(x + 1)}}} \right]_{\text{1}}^{\text{3}} \hfill \\{\text{ = }}\left[ {{\text{log}}\dfrac{{{\text{(x + 1)}}}}{{\text{x}}}\dfrac{{\text{1}}}{{\text{x}}}} \right]_{\text{1}}^{\text{3}} \hfill \\{\text{ = log}}\left( {\dfrac{{\text{4}}}{{\text{3}}}} \right){\text{ - }}\dfrac{{\text{1}}}{{\text{3}}}{\text{ - log}}\left( {\dfrac{{\text{2}}}{{\text{1}}}} \right){\text{ + 1}} \hfill \\{\text{ = log4 - log3 - log2 + }}\dfrac{{\text{2}}}{{\text{3}}} \hfill \\{\text{ = log}}\left( {\dfrac{{\text{2}}}{{\text{3}}}} \right){\text{ + }}\dfrac{{\text{2}}}{{\text{3}}} \hfill \\ \end{align}

इसलिए परिणाम साबित होता है।

35. $\int_{\text{0}}^{\text{1}} {{\text{x }}} {{\text{e}}^{\text{x}}}{\text{dx = 1}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{0}}^{\text{1}} {{\text{x }}} {{\text{e}}^{\text{x}}}{\text{dx}}$

\begin{align}{\text{ = x}}\int_{\text{0}}^{\text{1}} {{{\text{e}}^{\text{x}}}} {\text{\;dx - }}\int_{\text{0}}^{\text{1}} {\left\{ {\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{(x)}}\int {{{\text{e}}^{\text{x}}}} {\text{\;dx}}} \right\}} {\text{dx}} \hfill \\{\text{ = }}\left[ {{\text{x}}{{\text{e}}^{\text{x}}}} \right]_{\text{0}}^{\text{1}}{\text{ - }}\left[ {{{\text{e}}^{\text{x}}}} \right]_{\text{0}}^{\text{1}} \hfill \\ \end{align}

\begin{align}{\text{ = e - e + 1}} \hfill \\{\text{ = 1}} \hfill \\ \end{align}

इसलिए परिणाम साबित होता है।

36. $\int_{{\text{ - 1}}}^{\text{1}} {{{\text{x}}^{{\text{17}}}}} {\text{co}}{{\text{s}}^{\text{4}}}{\text{x dx = 0}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{{\text{ - 1}}}^{\text{1}} {{{\text{x}}^{{\text{17}}}}} {\text{co}}{{\text{s}}^{\text{4}}}{\text{x dx}}$

माना ${\text{f(x) = }}{{\text{x}}^{{\text{17}}}}{\text{co}}{{\text{s}}^{\text{4}}}{\text{x}}$

इसलिए ${\text{f( - x) = ( - x}}{{\text{)}}^{{\text{17}}}}{\text{co}}{{\text{s}}^{\text{4}}}{\text{( - x) = - }}{{\text{x}}^{{\text{17}}}}{\text{co}}{{\text{s}}^{\text{4}}}{\text{x = - f(x)}}$

इसलिए ${\text{f(x)}}$ विषम कार्य है

हमे पता है की अगर ${\text{f(x)}}$ विषम कार्य है तो $\int_{{\text{ - a}}}^{\text{a}} {\text{f}} {\text{(x)dx = 0}}$

इसलिए ${\text{I = }}\int_{{\text{ - 1}}}^{\text{1}} {{{\text{x}}^{{\text{17}}}}} {\text{co}}{{\text{s}}^{\text{4}}}{\text{x dx = 0}}$

इसलिए परिणाम साबित होता है।

37. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{3}}}} {\text{x dx = }}\dfrac{{\text{2}}}{{\text{3}}}$

उत्तर: ${\text{I}}\;{\text{ = }}\;$ $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{3}}}} {\text{x dx}}$

\begin{align}{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{si}}{{\text{n}}^{\text{2}}}} {\text{x sinx dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {\left( {{\text{1 - co}}{{\text{s}}^{\text{2}}}{\text{x}}} \right)} {\text{sinx dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{sin}}} {\text{x dx - }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{x sinx dx}} \hfill \\{\text{ = [ - cosx]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}}{\text{ + }}\left[ {\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x}}}}{{\text{3}}}}\right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{2}}}} \hfill \\{\text{ = 1 + }}\dfrac{{\text{1}}}{{\text{3}}}{\text{[ - 1]}} \hfill \\{\text{ = 1 - }}\dfrac{{\text{1}}}{{\text{3}}} \hfill \\{\text{I = }}\dfrac{{\text{2}}}{{\text{3}}} \hfill \\ \end{align}

38. $\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\text{2}} {\text{ ta}}{{\text{n}}^{\text{3}}}{\text{x dx = 1 - log2}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\text{2}} {\text{ ta}}{{\text{n}}^{\text{3}}}{\text{x dx}}$

\begin{align}{\text{ = 2}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{ta}}{{\text{n}}^{\text{2}}}} {\text{x tanx dx}} \hfill \\{\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x - 1}}} \right)} {\text{tanx dx}} \hfill \\ \end{align}

${\text{ = }}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {\left( {{\text{se}}{{\text{c}}^{\text{2}}}{\text{x tanx}}} \right)} {\text{dx - 2}}\int_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} {{\text{tan}}} {\text{x dx}}$

\begin{align}{\text{ = 2}}\left[ {\dfrac{{{\text{ta}}{{\text{n}}^{\text{2}}}{\text{x}}}}{{\text{2}}}} \right]_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}}{\text{ + 2[log cosx]}}_{\text{0}}^{\dfrac{{{\pi }}}{{\text{4}}}} \hfill \\{\text{ = (1 - 0) + 2}}\left[ {{\text{log cos}}\dfrac{{{\pi }}}{{\text{4}}}{\text{ - log cos0}}} \right] \hfill \\{\text{ = 1 + 2}}\left[ {{\text{log}}\dfrac{{\text{1}}}{{\sqrt {\text{2}} }}{\text{ - log1}}} \right] \hfill \\{\text{I = 1 - log2 - 0 = 1 - log2}} \hfill \\ \end{align}

39. $\int_{\text{0}}^{\text{1}} {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{x dx = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 1}}$

उत्तर: ${\text{I = }}\int_{\text{0}}^{\text{1}} {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{x dx = }}\int_{\text{0}}^{\text{1}} {{\text{si}}{{\text{n}}^{{\text{ - 1}}}}} {\text{x}}{\text{.1 dx}}$

भिन्न विभागों के समाकलन करके

\begin{align}{\text{I = }}\left[ {{\text{x si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}}{\text{ - }}\int_{\text{0}}^{\text{1}} {\dfrac{{\text{1}}}{{\sqrt {\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)} }}} {\text{x dx}} \hfill \\{\text{ = }}\left[ {{\text{x si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{0}}^{\text{1}} {\dfrac{{{\text{( - 2x)}}}}{{\sqrt {\left( {{\text{1 - }}{{\text{x}}^{\text{2}}}} \right)} }}} {\text{dx}} \hfill \\ \end{align}

माना ${\text{1 - }}{{\text{x}}^{\text{2}}}{\text{ = t}}$

\begin{align}{\text{ - 2x dx = dt}} \hfill \\{\text{x = 0, t = 1 ; x = 1, t = 0}} \hfill \\{\text{I = }}\left[ {{\text{x si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}\int_{\text{1}}^{\text{0}} {\dfrac{{{\text{dt}}}}{{\sqrt {\text{t}} }}} \hfill \\{\text{ = }}\left[ {{\text{x si}}{{\text{n}}^{{\text{ - 1}}}}{\text{x}}} \right]_{\text{0}}^{\text{1}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{[2}}\sqrt {\text{t}} {\text{]}}_{\text{1}}^{\text{0}} \hfill \\{\text{ = si}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1) + [ - }}\sqrt {\text{1}} {\text{]}} \hfill \\ \end{align}

${\text{I = }}\dfrac{{{\pi }}}{{\text{2}}}{\text{ - 1}}$

इसलिए परिणाम साबित होता है।

40. $\int {\dfrac{{{\text{dx}}}}{{{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}}}}$ की सीमा के रूप मे $\int_{\text{0}}^{\text{1}} {{{\text{e}}^{{\text{2 - 3x}}}}} {\text{dx}}$ का मान ज्ञात कीजिए।

उत्तर: ${\text{I = }}$ $\int_{\text{0}}^{\text{1}} {{{\text{e}}^{{\text{2 - 3x}}}}} {\text{dx}}$

हमे पता है की $\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x) = (b - a)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(a) + f(a + h) + \ldots + f(a + (n - 1)h)]}}$

जहा ${\text{h = }}\dfrac{{{\text{b - a}}}}{{\text{n}}}{\text{, a = 0, b = 1, f(x) = }}{{\text{e}}^{{\text{2 - 3x}}}}$

\begin{align}{\text{h = }}\dfrac{{{\text{b - a}}}}{{\text{n}}}{\text{ }} \Rightarrow \;{\text{h = }}\dfrac{{\text{1}}}{{\text{n}}} \hfill \\{\text{ = }}\int_{\text{a}}^{\text{b}} {{{\text{e}}^{{\text{2 - 3x}}}}} {\text{ dx = (1 - 0)}}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[f(0) + f(0 + h) + \ldots + f(0 + (n - 1)h)]}} \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{e}}^{\text{2}}}{\text{ + }}{{\text{e}}^{{\text{2 - 3h}}}}{\text{ + \ldots + }}{{\text{e}}^{{\text{2 - 3(n - 1)h}}}}} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}{\text{[}}{{\text{e}}^{\text{2}}}{\text{1 + }}{{\text{e}}^{{\text{ - 3h}}}}{\text{ + \ldots + }}{{\text{e}}^{{\text{ - 3(n - 1)h}}}}] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{e}}^{\text{2}}}\left\{ {\dfrac{{{\text{1 - }}{{\left( {{{\text{e}}^{{\text{ - 3h}}}}} \right)}^{\text{n}}}}}{{{\text{1 - }}{{\text{e}}^{{\text{ - 3h}}}}}}} \right\}} \right] \hfill \\{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {{{\text{e}}^{\text{2}}}\left\{ {\dfrac{{{\text{1 - }}{{\text{e}}^{\dfrac{{{\text{ - 3n}}}}{{\text{n}}}}}}}{{{\text{1 - }}{{\text{e}}^{{\text{ - 3n}}}}}}} \right\}} \right]{\text{ = }}\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{1}}}{{\text{n}}}\left[ {\dfrac{{{{\text{e}}^{\text{2}}}\left( {{\text{1 - }}{{\text{e}}^{{\text{ - 3}}}}} \right)}}{{{\text{1 - }}{{\text{e}}^{\dfrac{{\text{3}}}{{\text{n}}}}}}}} \right] \hfill \\{\text{ = }}{{\text{e}}^{\text{2}}}\left( {{{\text{e}}^{{\text{ - 3}}}}{\text{ - 1}}} \right)\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{3}}}} \right)\left[ {\dfrac{{\dfrac{{{\text{ -3}}}}{{\text{n}}}}}{{{{\text{e}}^{\dfrac{{{\text{ - 3}}}}{{\text{n}}}}}{\text{ - 1}}}}} \right] \hfill \\ \end{align}

\begin{align}{\text{ = }}\dfrac{{{{\text{e}}^{\text{2}}}}}{{\text{3}}}\left( {{{\text{e}}^{{\text{ - 3}}}}{\text{ - 1}}} \right)\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \left[ {\dfrac{{{\text{ - }}\dfrac{{\text{3}}}{{\text{n}}}}}{{{{\text{e}}^{\dfrac{{{\text{ - 3}}}}{{\text{n}}}}}{\text{ - 1}}}}} \right] \hfill \\{\text{ = }}\; - \dfrac{{{{\text{e}}^{\text{2}}}{\text{(}}{{\text{e}}^{{\text{ - 3}}}}{\text{ - 1)}}}}{{\text{3}}}\;\;\,\;\,\,\;\;\;[\because \;\mathop {{\text{lim}}}\limits_{{\text{n}} \to \infty } \dfrac{{\text{x}}}{{{{\text{e}}^{\dfrac{{\text{3}}}{{\text{n}}}{\text{ - 1}}}}}}{\text{ = 1}}] \hfill \\ {\text{ = - }}\dfrac{{{{\text{e}}^{\text{2}}}\left( {{{\text{e}}^{{\text{ - 3}}}}{\text{ - 1}}} \right)}}{{\text{3}}}{\text{ = }}\dfrac{{{\text{ - }}{{\text{e}}^{{\text{ - 1}}}}{\text{ + }}{{\text{e}}^{\text{2}}}}}{{\text{3}}} \hfill \\\because {\text{ I = }}\dfrac{{\text{1}}}{{\text{3}}}\left( {{{\text{e}}^{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{e}}}} \right) \hfill \\ \end{align}

### 41 से 44 तक के प्रश्नों मे सही उत्तर का चयन कीजिए:

41. $\int {\dfrac{{\text{1}}}{{{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}}}} {\text{dx}}$ बराबर है।

(a) ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{\text{e}}^{\text{x}}}} \right){\text{ + C}}$

(b) ${\text{ta}}{{\text{n}}^{{\text{ - 1}}}}\left( {{{\text{e}}^{{\text{ - x}}}}} \right){\text{ + C}}$

(c) ${\text{log}}\left( {{{\text{e}}^{\text{x}}}{\text{ - }}{{\text{e}}^{{\text{ - x}}}}} \right){\text{ + C}}$

(d) ${\text{log}}\left( {{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}} \right){\text{ + C}}$

उत्तर: $\int {\dfrac{{\text{1}}}{{{{\text{e}}^{\text{x}}}{\text{ + }}{{\text{e}}^{{\text{ - x}}}}}}} {\text{dx}}$

${\text{ = }}\int {\dfrac{{{{\text{e}}^{\text{x}}}}}{{{{\text{e}}^{{\text{2x}}}}{\text{ + 1}}}}} {\text{ dx}}$

माना ${{\text{e}}^{\text{x}}}{\text{ = t}}$

\begin{align}{{\text{e}}^{\text{x}}}{\text{ dx = dt}} \hfill \\{\text{I = }}\int {\dfrac{{\text{1}}}{{{\text{1 + }}{{\text{t}}^{\text{2}}}}}} {\text{ dt}} \hfill\\{\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{t + C}} \hfill \\{\text{I = ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{ }}{{\text{e}}^{\text{x}}}{\text{ + C}} \hfill \\ \end{align}

अतः विकल्प (a) सही है।

42. $\int {\dfrac{{{\text{cos2x}}}}{{{{{\text{(sinx + cosx)}}}^{\text{2}}}}}} {\text{dx}}$ बराबर है।

(a) $\dfrac{{{\text{ - 1}}}}{{{\text{sinx + cosx}}}}{\text{ + C}}$

(b) ${\text{log}}\mid {\text{sinx + cosx}}\mid {\text{ + C}}$

(c) ${\text{log|sinx - cosx| + C}}$

(d) $\dfrac{{\text{1}}}{{{{{\text{(sinx + cosx)}}}^{\text{2}}}}}$

उत्तर: ${\text{I}}\;{\text{ = }}$ $\int {\dfrac{{{\text{cos2x}}}}{{{{{\text{(sinx + cosx)}}}^{\text{2}}}}}} {\text{dx}}$

\begin{align}{\text{ = }}\int {\dfrac{{{\text{co}}{{\text{s}}^{\text{2}}}{\text{x - si}}{{\text{n}}^{\text{2}}}{\text{x}}}}{{{{{\text{(sinx + cosx)}}}^{\text{2}}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(sinx + cosx)(sinx - cosx)}}}}{{{{{\text{(sinx + cosx)}}}^{\text{2}}}}}} {\text{ dx}} \hfill \\{\text{ = }}\int {\dfrac{{{\text{(cosx - sinx)}}}}{{{\text{(sinx + cosx)}}}}} {\text{ dx}} \hfill \\ \end{align}

माना ${\text{sinx - cosx = t}}$

\begin{align}{\text{(sinx + cosx)dx = dt}} \hfill \\{\text{I = }}\int {\dfrac{{\text{1}}}{{\text{t}}}} {\text{ dt}} \hfill \\ \end{align}

\begin{align}{\text{ = log|t| + C}} \hfill \\{\text{ = log|sinx - cosx| + C}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

43. यदि ${\text{f(a + b - x) = f(x)}}$ तो $\int_{\text{a}}^{\text{b}} {\text{x}} {\text{ f(x) dx}}$ बराबर है।

(a) $\dfrac{{{\text{a + b}}}}{{\text{2}}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(b - x)dx}}$

(b) $\dfrac{{{\text{a + b}}}}{{\text{2}}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(b + x)dx}}$

(c) $\dfrac{{{\text{b - a}}}}{{\text{2}}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x)dx}}$

(d) $\dfrac{{{\text{a + b}}}}{{\text{2}}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x)dx}}$

उत्तर: ${\text{f(a + b - x) = f(x)}}$

माना ${\text{I}}\;{\text{ = }}$ $\int_{\text{a}}^{\text{b}} {\text{x}} {\text{ f(x) dx}}$ ………………..(i)

\begin{align}{\text{ = }}\int_{\text{a}}^{\text{b}} {{\text{(a + b - x)}}} {\text{f(a + b - x)dx}} \hfill \\\left\{ {\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x)dx}}} \right.\left. {{\text{ = }}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(a + b - x)dx}}} \right\} \hfill \\{\text{ = }}\int_{\text{a}}^{\text{b}} {{\text{(a + b - x)}}} {\text{f(x)dx}} \hfill \\{\text{I = (a + b)}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x) dx - 1}} \hfill \\ \end{align}

\begin{align}{\text{2I = (a + b)}}\int_{\text{a}}^{\text{b}} {\text{f}} {\text{(x) dx}} \hfill \\{\text{I = }}\dfrac{{{\text{(a + b)}}}}{{\text{2}}}\int_{\text{a}}^{\text{b}} {\text{f}}{\text{(x) dx}} \hfill \\ \end{align}

अतः विकल्प (d) सही है।

44. $\int_{\text{0}}^{\text{1}} {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{2x - 1}}}}{{{\text{1 + x - }}{{\text{x}}^{\text{2}}}}}} \right){\text{dx}}$ का मान है।

(a) ${\text{1}}$

(b) ${\text{0}}$

(c) ${\text{ - 1}}$

(d) $\dfrac{{{\pi }}}{{\text{4}}}$

उत्तर: ${\text{I}}\;{\text{ = }}$ $\int_{\text{0}}^{\text{1}} {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{2x - 1}}}}{{{\text{1 + x - }}{{\text{x}}^{\text{2}}}}}} \right){\text{dx}}$

${\text{I = }}\int_{\text{0}}^{\text{1}} {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}} \left( {\dfrac{{{\text{x - (1 - x)}}}}{{{\text{1 + x(1 - x)}}}}} \right){\text{ dx}}$

${\text{I = }}\int_{\text{0}}^{\text{1}} {\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x) - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x)}}} \right]} {\text{ dx}}$ ……………..(1)

${\text{I = }}\int_{\text{0}}^{\text{1}} {\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x) - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - 1 + x)}}} \right]} {\text{ dx}}$

${\text{I = }}\int_{\text{0}}^{\text{1}} {\left[ {{\text{ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x) - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x)}}} \right]} {\text{dx}}$ ………………….(2)

(1)+(2)

${\text{2I = }}\int_{\text{0}}^{\text{1}} {{\text{[ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x) + ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x) - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(1 - x) - ta}}{{\text{n}}^{{\text{ - 1}}}}{\text{(x)] dx}}}$

\begin{align} {\text{2I}}\;{\text{ = }}\;{\text{0}} \hfill \\{\text{I}}\;{\text{ = }}\;{\text{0}} \hfill \\ \end{align}

अतः विकल्प (b) सही है।

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