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

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 - FREE PDF Download

NCERT Solutions for Class 12 Maths Exercise 7.4 PDF helps students understand the detailed method of solving the sums of integration. Integration is one of the most important topics covered in the Maths syllabus of Class 12. Vedantu provides detailed explanations and step-by-step solutions for integrating specific functions using various methods such as substitution, trigonometric identities, integration by parts, and partial fractions. Class 12 Ex 7.4 is crucial for students to understand and apply different integration techniques essential for solving complex calculus problems. By practicing the 25 questions in this exercise, students can build a solid foundation in indefinite integrals, a key concept in calculus that is frequently tested in board exams.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.4 Class 12 | Vedantu
3. Access PDF for Maths NCERT Chapter 7 Integrals Exercise 7.4 Class 12
4. Class 12 Maths Chapter 7: Exercises Breakdown
5. CBSE Class 12 Maths Chapter 7 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 7 Exercise 7.4 Class 12 | Vedantu

• Integrals are a fundamental concept in calculus, and they can be used to find the area under a curve, the volume of a solid, and many other things.

• Exercise 7.4 of NCERT Class 12 Maths focuses on the integration of some particular functions, which are commonly needed in calculus.

• The process of finding the integral of a function is called integration. It's the opposite of differentiation.

• Adding an arbitrary constant C to an indefinite integral gives another indefinite integral of the same function.

• The value of a definite integral is independent of the variable of integration.

• The integral of a constant function over an interval is equal to the product of the constant and the length of the interval.

• The integral of a function over an interval can be split into the sum of the integrals over subintervals of that interval.

• Exercise 7.4 contains 25 fully solved questions, offering extensive practice to ensure proficiency.

Competitive Exams after 12th Science

## Access PDF for Maths NCERT Chapter 7 Integrals Exercise 7.4 Class 12

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.

## Conclusion

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.4 by Vedantu covers the integration of specific functions using various methods such as substitution, trigonometric identities, integration by parts, and partial fractions. This exercise is crucial for mastering the calculation of indefinite integrals, which is a fundamental concept in calculus. Students should focus on understanding these integration techniques as they form the basis for solving complex problems and are frequently tested in exams.

## Class 12 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.1 22 Questions & Solutions (21 Short Answers, 1 MCQs) Exercise 7.2 39 Questions & Solutions (37 Short Answers, 2 MCQs) Exercise 7.3 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.5 23 Questions & Solutions (21 Short Answers, 2 MCQs) Exercise 7.6 24 Questions & Solutions (22 Short Answers, 2 MCQs) Exercise 7.7 11 Questions & Solutions (9 Short Answers, 2 MCQs) Exercise 7.8 22 Questions & Solutions (20 Short Answers, 2 MCQs) Exercise 7.9 10 Questions & Solutions (8 Short Answers, 2 MCQs) Exercise 7.10 21 Questions & Solutions (19 Short Answers, 2 MCQs)

## Chapter-Specific NCERT Solutions for Class 12 Maths

The chapter-wise NCERT Solutions for Class 12 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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

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.

9. What are the key concepts covered in Ex 7.4 Class 12 Maths NCERT Solutions?

Exercise 7.4 focuses on the application of integration by parts, integration using partial fractions, and some specific integrals involving trigonometric functions.

10. What are some common mistakes to avoid while solving Exercise 7.4 Class 12 Maths?

Common mistakes include:

• Incorrectly applying the integration by parts formula

• Choosing the wrong function for 'u' in integration by parts

• Errors in decomposing rational functions into partial fractions

• Misapplication of trigonometric identities

11. What are partial fractions and how are they used in Class 12 Maths 7.4?

Partial fractions are a way of expressing a rational function as the sum of simpler fractions. In Exercise 7.4, they are used to decompose complex rational integrals into simpler fractions that can be easily integrated.