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# NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.4

Last updated date: 09th Apr 2024
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## NCERT Solutions for Class 12 Maths Chapter 7 (Ex 7.4)

Integration is one of the most important topics covered in the Maths syllabus of Class 12. So, the highly experienced teachers at Vedantu prepared the Exercise 7.4 Class 12 Maths NCERT Solution PDF to help students understand the detailed method of solving the sums of integration. Students can download Class 12 Maths Chapter 7 Exercise 7.4 Solutions PDF from Vedantu for free. These solutions will help students grasp all the major concepts covered in the chapter. This, in turn, will help students to score the highest marks in their exams.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Maths Chapter Name: Chapter 7 - Integrals Exercise: Exercise - 7.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
Competitive Exams after 12th Science

Exercise 7.4

1. Find the integration of$\dfrac{{3{x^2}}}{{{x^6} + 1}}$.

Ans: Let${x^3} = t$,

Now, differentiate both sides,

$3{x^2}dx = dt$

$\displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx} = \displaystyle \int {\dfrac{{3{x^2}}}{{{{\left( {{x^3}} \right)}^2} + 1}}dx}$

$\displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx} = \displaystyle \int {\dfrac{1}{{{t^2} + 1}}dt}$

$\displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx} = {\tan ^{ - 1}}t + C$

$\displaystyle \int {\dfrac{{3{x^2}}}{{{x^6} + 1}}dx} = {\tan ^{ - 1}}\left( {{x^3}} \right) + C$

Where C is an arbitrary constant.

2. Find the integration of $\dfrac{1}{{\sqrt {1 + 4{x^2}} }}$.

Ans: Let $2x = t$,

Now, differentiate both sides,

$2dx = dt$

$dx = \dfrac{{dt}}{2}$

$\displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {1 + {{\left( {2x} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}\left( {\dfrac{{dt}}{2}} \right)}$

$\displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \dfrac{1}{2}\left( {\log |t + \sqrt {1 + {t^2}} |} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \log |x + \sqrt {{x^2} + {a^2}} |} } \right]$

$\displaystyle \int {\dfrac{1}{{\sqrt {1 + 4{x^2}} }}dx} = \dfrac{1}{2}\log |2x + \sqrt {4{x^2} + 1} | + C$

Where C is an arbitrary constant.

3. Find the integration of $\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}$.

Ans: Let $2 - x = t$,

Now, differentiate both sides,

$0 - dx = dt$

$dx = - dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + 1} }}\left( { - dt} \right)}$

$\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx} = - \displaystyle \int {\dfrac{1}{{\sqrt {1 + {t^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx} = - \left( {\log |t + \sqrt {1 + {t^2}} |} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \log |x + \sqrt {{x^2} + {a^2}} |} } \right]$

$\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx} = - \log |\left( {2 - x} \right) + \sqrt {{{\left( {2 - x} \right)}^2} + 1} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}dx} = \log \left( {\dfrac{1}{{\left( {2 - x} \right) + \sqrt {{x^2} - 4x + 5} }}} \right) + C$

Where C is an arbitrary constant.

4. Find the integration of $\dfrac{1}{{\sqrt {9 - 25{x^2}} }}$.

Ans: Let $5x = t$,

Now, differentiate both sides,

$5dx = dt$

$dx = \dfrac{{dt}}{5}$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( {5x} \right)}^2}} }}} dx$

$\displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}} \left( {\dfrac{{dt}}{5}} \right)$

$\displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}}$

$\displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{t}{3}} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}dx} = {{\sin }^{ - 1}}\dfrac{x}{a}} \right]$

$\displaystyle \int {\dfrac{1}{{\sqrt {9 - 25{x^2}} }}} dx = \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{{5x}}{3}} \right) + C$

Where C is an arbitrary constant.

5. Find the integration of $\dfrac{{3x}}{{1 + 2{x^4}}}$.

Ans: Let $\sqrt 2 {x^2} = t$,

Now, differentiate both sides,

$2\sqrt 2 xdx = dt$

$xdx = \dfrac{{dt}}{{2\sqrt 2 }}$

Now,

$\displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx} = \displaystyle \int {\dfrac{{3x}}{{1 + {{\left( {\sqrt 2 {x^2}} \right)}^2}}}dx}$

$\displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx} = \displaystyle \int {\dfrac{3}{{1 + {t^2}}}\left( {\dfrac{{dt}}{{2\sqrt 2 }}} \right)}$

$\displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx} = \dfrac{3}{{2\sqrt 2 }}\displaystyle \int {\dfrac{1}{{1 + {t^2}}}dt}$

$\displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx} = \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}t + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{1 + {x^2}}}dx} = {{\tan }^{ - 1}}x + C} \right]$

$\displaystyle \int {\dfrac{{3x}}{{1 + 2{x^4}}}dx} = \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + C$

Where C is an arbitrary constant.

6. Find the integration of $\dfrac{{{x^2}}}{{1 - {x^6}}}$.

Ans: Let ${x^3} = t$,

Now, differentiate both sides,

$3{x^2}dx = dt$

${x^2}dx = \dfrac{{dt}}{3}$

Now,

$\displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx = } \displaystyle \int {\dfrac{{{x^2}}}{{1 - {{\left( {{x^3}} \right)}^2}}}dx}$

$\displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx} = \displaystyle \int {\dfrac{1}{{1 - {t^2}}}\left( {\dfrac{{dt}}{3}} \right)}$

$\displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx} = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{1 - {t^2}}}dt}$

$\displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx} = \dfrac{1}{3}\left( {\dfrac{1}{2}\log |\dfrac{{1 + t}}{{1 - t}}|} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{{a^2} - {x^2}}}dx} = \dfrac{1}{{2a}}\log |\dfrac{{a + x}}{{a - x}}| + C} \right]$

$\displaystyle \int {\dfrac{{{x^2}}}{{1 - {x^6}}}dx} = \dfrac{1}{3}\left( {\dfrac{1}{2}\log |\dfrac{{1 + {x^3}}}{{1 - {x^3}}}|} \right) + C$

Where C is an arbitrary constant.

7. Find the integration of $\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}$.

Ans: $\displaystyle \int {\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx} = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} - \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{ }}\left[ {eq.1} \right]$

For $\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx}$,

Let ${x^2} - 1 = t$,

Now differentiate both sides,

$2xdx = dt$

$xdx = \dfrac{{dt}}{2}$

Now,

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt t }}\left( {\dfrac{{dt}}{2}} \right)}$

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\displaystyle \int {{t^{ - \dfrac{1}{2}}}dt}$

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right) + C$

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\left( {2\sqrt t } \right) + C$

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \sqrt t + C$

$\displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} = \sqrt {{x^2} - 1} + C$

Now, from equation $1$,

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

$\displaystyle \int {\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx} = \sqrt {{x^2} - 1} - \log |x + \sqrt {{x^2} - 1} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - {a^2}} }}dx} = \log |x + \sqrt {{x^2} - {a^2}} |} \right]$

Where C is an arbitrary constant.

8. Find the integration of $\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}$.

Ans: Let ${x^3} = t$,

Now, differentiate both sides,

$3{x^2}dx = dt$

${x^2}dx = \dfrac{{dt}}{3}$

Now,

$\displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} = \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{{\left( {{x^3}} \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}\left( {\dfrac{{dt}}{3}} \right)}$

$\displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dt}$

$\displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} = \dfrac{1}{3}\log |t + \sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = } \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$\displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} = \dfrac{1}{3}\log |{x^3} + \sqrt {{x^6} + {a^6}} | + C$

Where C is an arbitrary constant.

9. Find the integration of $\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}$.

Ans: Let $\tan x = t$,

Now, differentiate both sides,

${\sec ^2}xdx = dt$

Now,

$\displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\left( {\tan x} \right)}^2} + {{\left( 2 \right)}^2}} }}} dx$

$\displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \displaystyle \int {\dfrac{{dt}}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }}}$

$\displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \log |t + \sqrt {{t^2} + {{\left( 2 \right)}^2}} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx} = \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$\displaystyle \int {\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}} dx = \log |\tan x + \sqrt {{{\tan }^2}x + 4} | + C$

Where C is an arbitrary constant.

10. Find the integration of $\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}$.

Ans: Let $x + 1 = t$,

Now, differentiate both sides,

$dx = dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( 1 \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {{\left( 1 \right)}^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |t + \sqrt {{t^2} + 1} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx} = \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 1} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}} dx = \log |x + 1 + \sqrt {{x^2} + 2x + 2} | + C$

Where C is an arbitrary constant.

11. Find the integration of $\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}$.

Ans: Let $3x + 1 = t$,

Now, differentiate both sides,

$3dx = dt$

$dx = \dfrac{{dt}}{3}$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 1 + 4} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}\left( {\dfrac{{dt}}{3}} \right)}$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \dfrac{1}{3}\log |t + \sqrt {{t^2} + {2^2}} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx} = \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {9{x^2} + 6x + 5} | + C$

Where C is an arbitrary constant.

12. Find the integration of $\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}$.

Ans:

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x} \right)} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x + 9 - 9} \right)} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {7 + 9 - {{\left( {x + 3} \right)}^3}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {16 - {{\left( {x + 3} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}$

Let,

$x + 3 = t$

$dx = dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {t^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = {\sin ^{ - 1}}\left( {\dfrac{t}{4}} \right) + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = {\sin ^{ - 1}}\left( {\dfrac{{x + 3}}{4}} \right) + C$

Where C is an arbitrary constant.

13. Find the integration of $\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}$.

Ans:

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + 2} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4} + 2} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \left( {\dfrac{1}{4}} \right)} }}} dx$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

Now, let

$x - \dfrac{3}{2} = t$

$dx = dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dt}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C$

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

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{x^2} - 3x + 2} | + C$

Where C is an arbitrary constant.

14. Find the integration of $\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}$.

Ans: Let $x - \dfrac{3}{2} = t$

Now, differentiate both sides,

$dx = dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4}} \right)} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {8 + \dfrac{9}{4} - \left( {{x^2} - 3x + \dfrac{9}{4}} \right)} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {\left( {\dfrac{{41}}{4}} \right) - {{\left( {{x^2} - \dfrac{3}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {t^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = {\sin ^{ - 1}}\dfrac{t}{{\dfrac{{\sqrt {41} }}{2}}} + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = {\sin ^{ - 1}}\dfrac{{2\left( {x - \dfrac{3}{2}} \right)}}{{\sqrt {41} }} + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} = {\sin ^{ - 1}}\dfrac{{2x - 3}}{{\sqrt {41} }} + C$

Where C is an arbitrary constant.

15. Find the integration of $\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}$.

Ans:

$\left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + ab$

$\left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + \dfrac{{{{\left( {a + b} \right)}^2}}}{4} - \dfrac{{{{\left( {a + b} \right)}^2}}}{4} + ab$

$\left( {x - a} \right)\left( {x - b} \right) = {\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{4}$

Now, let

$x - \left( {\dfrac{{a + b}}{2}} \right) = t$

$dx = dt$

Now,

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]}^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} = \log |t + \sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} = \log |\left\{ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right\} + \sqrt {\left( {x - a} \right)\left( {x - b} \right)} | + C$

Where C is an arbitrary constant.

16. Find the integration of $\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}$.

Ans: Let $2{x^2} + x - 3 = t$,

Now, differentiate both sides,

$\left( {4x + 1} \right)dx = dt$

Now,

$\displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt t }}} dt$

$\displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = \dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C$

$\displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = 2\sqrt t + C$

$\displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = 2\sqrt {2{x^2} + x - 3} + C$

Where C is an arbitrary constant.

17. Find the integration of $\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}$.

Ans:

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} + \displaystyle \int {\dfrac{2}{{\sqrt {{x^2} - 1} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{ }}\left[ {eq.1} \right]$

Now, for $\dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx}$, let ${x^2} - 1 = t$

Now, differentiate both sides,

$2xdx = dt$

Now, in equation $1$,

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{dt}}{{\sqrt t }}} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \dfrac{1}{2}\left( {2\sqrt t } \right) + 2\log |x + \sqrt {{x^2} - 1} | + C$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} = \sqrt {{x^2} - 1} + 2\log |x + \sqrt {{x^2} - 1} | + C$

Where C is an arbitrary constant.

18. Find the integration of $\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}$.

Ans:

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \displaystyle \int {\dfrac{{5x}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2 - 2}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx - \dfrac{5}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} } - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} {\text{ }}\left[ {eq.1} \right]$

Now, for$\dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx}$

Let $3{x^2} + 2x + 1 = t$,

Now, differentiate both sides,

$\left( {6x + 2} \right)dx = dt$

Now, in eq. $1$,

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x} \right) + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x + \dfrac{1}{9} - \dfrac{1}{9}} \right) + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} - \dfrac{1}{3} + 1}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} + \dfrac{2}{3}}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{9}\displaystyle \int {\dfrac{1}{{{{\left( {x + \dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2}}}dx}$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\log |t| - \dfrac{{11}}{9}\left( {\dfrac{3}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C$

$\displaystyle \int {\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx} = \dfrac{5}{6}\log |3{x^2} + 2x + 1| - \dfrac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C$

Where C is an arbitrary constant.

19. Find the integration of $\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}$.

Ans:

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = \displaystyle \int {\dfrac{{6x}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{{2x - 9 + 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + \displaystyle \int {\dfrac{{27}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} {\text{ }}\left[ {eq.1} \right]$

Now for $3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx}$,

Let ${x^2} - 9x + 20 = t$,

Differentiate both sides,

$\left( {2x - 9} \right)dx = dt$

Now, from eq. $1$,

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + \dfrac{{81}}{4} - \dfrac{{81}}{4} + 20} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\left( {2\sqrt t } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C$

$\displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} = 3\left( {2\sqrt {{x^2} - 9x + 20} } \right) + 34\log |\left( {x - \dfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} | + C$

Where C is an arbitrary constant.

20. Find the integration of $\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}$.

Ans:

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = \displaystyle \int {\dfrac{x}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{{ - 2x}}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x - 4}}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 2} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} {\text{ }}\left[ {eq.1} \right]$

Now, for $- \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx}$,

Let $4x - {x^2} = t$,

Now, differentiate both sides,

$\left( {4 - 2x} \right)dx = dt$

Now, from eq. $1$,

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt { - \left( {{x^2} - 4x + 4 - 4} \right)} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \dfrac{1}{2}\left( {2\sqrt t } \right) + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} = - \sqrt {4x - {x^2}} + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C$

Where C is an arbitrary constant.

21. Find the integration of $\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}$.

Ans:

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2 - 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx - } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} {\text{ [eq}}{\text{.1]}}$

Now, for $\dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx}$

Let ${x^2} + 2x + 3 = t$,

Now, differentiate both sides,

$\left( {2x + 2} \right)dx = dt$

Now, from eq. $1$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 1 + 2} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} = \dfrac{1}{2}\left( {2\sqrt t } \right) + \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 2} | + C$

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

Where C is an arbitrary constant.

22. Find the integration of $\dfrac{{x + 3}}{{{x^2} - 2x - 5}}$.

Ans:

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \displaystyle \int {\dfrac{x}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2 + 2}}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx} + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx} {\text{ }}\left[ {eq.1} \right]$

Now, for $\dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx}$

Let ${x^2} - 2x - 5 = t$,

Now, differentiate both sides,

$\left( {2x - 2} \right)dx = dt$

Now, from equation $1$,

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx} + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt} + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x + 1 - 1 - 5}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt} + 4\displaystyle \int {\dfrac{1}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}dx}$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\log |t| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C$

$\displaystyle \int {\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx} = \dfrac{1}{2}\log |{x^2} - 2x - 5| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C$

Where C is an arbitrary constant.

23. Find the integration of $\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}$.

Ans:

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \displaystyle \int {\dfrac{{5x}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4 - 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx} {\text{ }}\left[ {eq.1} \right]$

Now, for $\dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}$,

Let ${x^2} + 4x + 10 = t$,

Now, differentiate both sides,

$\left( {2x + 4} \right)dx = dt$

Now, from eq. $1$,

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 4 + 6} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = \dfrac{5}{2}\left( {2\sqrt t } \right) - 7\log |\left( {x + 2} \right) + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} | + C$

$\displaystyle \int {\dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} = 5\sqrt {{x^2} + 4x + 10} - 7\log |\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 10} | + C$

Where C is an arbitrary constant.

24. The integration $\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}$is equals to:

A. $x{\tan ^{ - 1}}\left( {x + 1} \right) + C$

B. ${\tan ^{ - 1}}\left( {x + 1} \right) + C$

C. $\left( {x + 1} \right){\tan ^{ - 1}}\left( x \right) + C$

D. ${\tan ^{ - 1}}\left( x \right) + C$

Ans:

$\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}} = \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 1 + 1}}}$

$\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}} = \displaystyle \int {\dfrac{{dx}}{{{{\left( {x + 1} \right)}^1} + 1}}}$

$\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}} = \dfrac{1}{1}{\tan ^{ - 1}}\left( {\dfrac{{x + 1}}{1}} \right) + C$

$\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}} = {\tan ^{ - 1}}\left( {x + 1} \right) + C$

Thus, the correct answer is B.

25. The integration $\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$is equals to:

A. $\dfrac{1}{9}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C$

B. $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C$

C. $\dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C$

D. $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{9}} \right) + C$

Ans:

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \displaystyle \int {\dfrac{{dx}}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x} \right)} }}}$

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x + \dfrac{{81}}{{64}} - \dfrac{{81}}{{64}}} \right)} }}dx}$

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left[ {{{\left( {x - \dfrac{9}{8}} \right)}^2} - {{\left( {\dfrac{9}{8}} \right)}^2}} \right]} }}dx}$

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{9}{8}} \right)}^2} - {{\left( {x - \dfrac{9}{8}} \right)}^2}} }}dx}$

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{x - \dfrac{9}{8}}}{{\dfrac{9}{8}}}} \right) + C$

$\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}} = \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C$

Thus, the correct answer is B.

Integral is a tricky chapter, and students will get a good understanding of the types of sums covered in integration by going through the Ex 7.4 Class 12 NCERT Solutions PDF available on Vedantu. However, before going through Class 12 Maths NCERT Solutions Chapter 7 Exercise 7.4, there are some concepts that students should be familiar with. In this section, we will talk about those important concepts.

Class 12 Maths Ex 7.4 Solutions, cover solutions to all the sums given in this exercise. In this chapter you will learn that integration can be explained as the method for finding out the areas of the two-dimensional region and computing volumes of three-dimensional objects.

This is why if an individual wants to find out the integral of a function with respect to x, then he or she needs to find the area to the x-axis from the curve. There are many experts who also refer to integrals as antiderivatives. This is because integration is the reverse process of differentiation.

There are also different types of integrals. Students will get to know about the two types of integrals in Maths NCERT Solutions Class 12 Chapter 7 Exercise 7.4. The two types of integrals are as follows.

Definite Integral

Definite integrals can be defined as the types of integrals that have definite limits, the upper limit and lower limit. While working on the NCERT Solution of Maths Class 12 Chapter 7 Exercise 7.4, students should remember that definite integrals are also known as Riemann integrals. These integrals can be expressed as:

∫(b upper limit and a lower limit) f (x) d (x).

Indefinite Integral

Indefinite integrals, on the other hand, are defined as integrals in which the upper and lower limits are not defined. This means that while solving the sums of Exercise 7.4 Class 12th Maths answers, you should represent indefinite integrals as:

∫ f (x) d (x) = F (x) + c

In this equation, c is the constant value.

It should be noted that the topic of indefinite integrals is more important for solving maths Exercise 7.4 Class 12 questions. Also, if integration is the inverse of differentiation, then

a / ax F (x) = f (x), then ∫ f (x) dx = F (x) + c

Indefinite integrals are also known as general integrals. These integrals differ in the constant term. From a geometrical perspective, an indefinite integral can be seen as a compilation of curves that are obtained through the translation of one of the curves parallel to itself download and upward along with the y-axis.

There are also several applications of integrals. In mathematics, the applications of integrals are:

• To find out the average value of a curve.

• To calculate the area under a curve.

• To find out the area between any two curves.

• To calculate the center of mass of the centroid of an area that has curved sides.

The applications of integrals in the subject of Physics are:

• To calculate the amount of thrust.

• To calculate the center of gravity.

• To find out the trajectory of a satellite at the time of placing it in an orbit.

• To find out the momentum and mass of inertia of vehicles.

• To calculate the momentum and mass of satellites.

• To find the center of mass.

• To calculator the velocity of a satellite at the time of placing it inside the orbit.

It should also be noted that integration denotes the summation of discrete data. The integral can also be calculated to find the functions that will describe the displacement, area, and volume that occurs because of a collection of small data. This value cannot be measured singularly.

In a broad sense, in the field of calculus, the idea of limit is used for geometry and algebra. The limits help in studying the result of points on a graph like how the points get closer to one another until the distance almost measures to be almost zero.

There are also two major types of calculus. These types of calculus are:

• Integral Calculus.

• Differential Calculus.

The concept of integration can also help in finding out the problem function when the derivatives are mentioned. Also, the concept can be used to find the area bounded by the graph of a function under various constraints.

These points are also important for the development of the concept of ‘Integral Calculus.’ Integral calculus consists of indefinite and definite integrals. Also, in calculus, the concept of differentiating a function and integrating a function is linked through a theorem, which is known as the Fundamental Theorem of Calculus.

### NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4

Concepts covered in the exercise:

• Methods of integration

• Integration by substitution

• Integration using trigonometric identities

• Integrals for some particular functions

• Integration by partial fractions

### Download Ex 7.4 Class 12 Maths NCERT Solutions PDF from Vedantu

Integral is an important chapter in mathematics. You can download and refer to the solved NCERT Solutions PDF from Vedantu.

### NCERT Solution Class 12 Maths of Chapter 7 All Exercises

Students should solve all the questions given in the other exercises. We at Vedantu provide solutions of all the exercises of chapter 7 class 12th. The PDF files of these solutions are available for free.

There are several benefits to downloading NCERT maths solutions, and some of those benefits are mentioned in the list below.

• All solutions are present in a PDF format.

• The PDF file can be downloaded at any time and anywhere on-the-go.

• All sums are solved by our subject-matter experts explaining every step clearly.

• The solutions are prepared according to the guidelines set by CBSE.

• Students can also enroll for live online classes at Vedantu.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.4

1. Where can I get Solved NCERT Solutions for Class 12 Maths Chapter 7?

Students can get solved NCERT Solutions for Class 12 Maths at the official site of Vedantu. It is an online learning platform that provides students with various solutions and notes for easy understanding. These solutions are made by our subject-matter experts who have years of experience in this similar field.

3. How many Chapters are included in the Class 12 Maths Syllabus?

It is advisable to the students to keep a clear knowledge of the NCERT Class 12 Syllabus before preparing for their board’s examination. Having a deep understanding of the syllabus will help them score better. Class 12 Maths syllabus includes 6 units along with the following topics:

• Unit 1 - Relations and Functions

Relations and functions

Inverse Trigonometry

• Unit 2 - Algebra

Matrices

Determinants

• Unit 3 - Calculus

Continuity and Differentiability

Applications of Derivatives

Integrals

Applications of the Integrals

Differential Equations

• Unit 4 - Vectors and Three-Dimensional Geometry

Vectors

Three-Dimensional Geometry

• Unit 5 - Linear Programming

• Unit 6 - Probability

Multiplications theorem on probability

4. Is R.S. Aggarwal a Good reference Book for Class 12 Maths?

Yes, R.S. Aggarwal is a good reference book for Class 12 Maths. The book provides detailed chapter-wise solutions in a very simple and understandable manner. Each chapter is provided with solved solutions which are categorised into easy, moderate and tough questions. However, students are also advised to refer to the NCERT textbook and sample papers along with the reference books to get a better understanding of the subject. This will help them in scoring good marks.

5. What topics does Exercise 7.4 of Chapter 7 of Class 12 Maths focus on?

Chapter 7 in Class 12 Maths is called Integrals. While the chapter includes details about various methods that can be used for solving sums of integration, Exercise 7.4 introduces the two types of integrals and the method specifically used for calculating the areas of the two-dimensional region and determining volumes of three-dimensional objects. Students shall be able to use the introduced methods in solving the sums given in Exercise 7.4.

6. How many questions are there in the NCERT Solutions for Exercise 7.4 of Chapter 7 of Class 12 Maths?

Chapter 7 of Class 12 Maths consists of 11 exercises in total. Exercise 7.4 in this chapter contains a total of 25 questions which have been answered step-by-step in the NCERT Solutions for Exercise 7.4 of Chapter 7 of Class 12 Maths. Students are advised to practise all the questions and solutions to strengthen their understanding of the concepts taught in this chapter.

7. Should I practise all the examples based on Exercise 7.4 of Chapter 7 of Class 12 Maths?

Examples are an important part of the syllabus of Class 12 Maths. All the examples are based on the subsequent exercises and practising them can help you solve questions in the exercise as well. All the students are advised to practise all the examples available in Exercise 7.4 of Chapter 7 of Class 12 Maths as they will help you get clarity about the methods and concepts. Questions based on the examples are often asked in the exam as well.

8. What is Exercise 7.4 of Chapter 7 of Class 12 Maths all about?

Exercise 7.4 of Chapter 7 of Class 12 Maths is about Integrals of Particular Functions. It is an important topic and students need to be well-versed in it to score good marks. For a detailed explanation or important questions of this exercise, you can also visit Vedantu where all the solutions curated by experts are available free of cost. These solutions are available on Vedantu’s website(vedantu.com) or mobile app.