NCERT Solutions for Class 12 Maths Chapter 7 - Integrals

Chapter 7 of the Class 12 Maths NCERT textbook deals with Integrals. This chapter covers various topics such as indefinite integrals, definite integrals, methods of integration, integration using trigonometric identities, integration by substitution, integration by parts, integration using partial fractions, and miscellaneous integrals.

The exercises in this chapter are designed to help students gain a comprehensive understanding of integrals and their various applications.

Exercise 7.1 This exercise introduces the concept of indefinite integrals and teaches students how to find them.

Exercise 7.2 This exercise deals with the properties of indefinite integrals and helps students understand how to use these properties to solve problems.

Exercise 7.3 This exercise covers the concept of definite integrals and teaches students how to evaluate them using different methods such as the limit definition of definite integrals and the properties of definite integrals.

Exercise 7.4 This exercise teaches students how to evaluate definite integrals using substitution.

Exercise 7.5 This exercise deals with integration by parts and teaches students how to solve problems using this method.

Exercise 7.6 This exercise covers the concept of partial fractions and helps students understand how to use them to evaluate integrals.

Exercise 7.7 This exercise teaches students how to evaluate integrals using trigonometric identities.

Exercise 7.8 This exercise covers the concept of integration using trigonometric substitutions and helps students understand how to use this method to solve problems.

Exercise 7.9 This exercise teaches students how to evaluate integrals using the method of differentiation under the integral sign.

Exercise 7.10 This exercise deals with the concept of improper integrals and teaches students how to evaluate them.

Exercise 7.11 This exercise covers various miscellaneous integrals and teaches students how to solve problems using different methods.

Miscellaneous Exercise: This exercise contains a collection of problems that require students to apply the concepts they have learned in the previous exercises.

In summary, this chapter covers a wide range of topics related to integrals, and the exercises are designed to help students develop a deep understanding of the concepts and their applications.

Access NCERT Solutions for Class 12 Mathematics  Chapter 7- Integrals

Exercise 7.1

1. Find an antiderivative (or integral) of the following functions by the method of inspection. sin 2x

Ans:  We use the method of inspection as follows:

$ \dfrac{d}{dx}\left( \cos 2x \right)=-2\sin 2x\Rightarrow -\dfrac{1}{2}\dfrac{d}{dx}\left( \cos 2x \right) $

$\therefore \sin 2x=\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x \right)$

Thus, the anti-derivative of sin $2x$is $-\dfrac{1}{2}\cos 2x$.

2. Find an antiderivative (or integral) of the following functions by the method of inspection. cos 3x

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( \sin 3x \right)=3\cos 3x\Rightarrow \dfrac{1}{3}\dfrac{d}{dx}\left( \sin 3x \right)$

 $ \therefore \cos 3x=\dfrac{d}{dx}\left( \dfrac{1}{3}\sin 3x \right)$

        Thus, the anti - derivative of cos $3x$is $\dfrac{1}{3}\sin 3x$.

3. Find an antiderivative (or integral) of the following functions by the method of inspection. $\mathbf{{{e}^{2x}}}$

Ans: We use the method of inspection as follows:

$ \dfrac{d}{dx}\left( {{e}^{2x}} \right)\Rightarrow 2{{e}^{2x}}=\dfrac{1}{2}\dfrac{d}{dx}\left( {{e}^{2x}} \right) $

$ \therefore {{e}^{2x}}=\dfrac{d}{dx}\left( \dfrac{1}{2}{{e}^{2x}} \right) $

Thus, the anti-derivative of ${{e}^{2x}}$is $\dfrac{1}{2}{{e}^{2x}}$.

4. Find an antiderivative (or integral) of the following functions by the method of inspection.$\mathbf{{{\left( ax+b \right)}^{2}}}$

Ans: We use the method of inspection as follows:

  $  \dfrac{d}{dx}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}} $ 

$ \Rightarrow {{\left( ax+b \right)}^{2}}=\dfrac{1}{3a}\dfrac{d}{dx}{{\left( ax+b \right)}^{3}} $ 

 $  \therefore {{\left( ax+b \right)}^{2}}=\dfrac{d}{dx}\left( \dfrac{1}{3a}{{\left( ax+b \right)}^{3}} \right)$

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\dfrac{1}{3a}{{\left( ax+b \right)}^{3}}$.

5. Find an antiderivative (or integral) of the following functions by the method of inspection. sin $\mathbf{2x-4{{e}^{3x}}}$

Ans: We use the method of inspection as follows:

$\dfrac{d}{dx}\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)=\left( \sin 2x-4{{e}^{3x}} \right)$

Thus, the anti-derivative of $\left( \sin 2x-4{{e}^{3x}} \right)$ is $\left( -\dfrac{1}{2}\cos 2x-\dfrac{4}{3}{{e}^{3x}} \right)$.

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

Ans: 

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

$ =4\int{{{e}^{3x}}dx}+\int{1}dx $

$ =4\left( \dfrac{{{e}^{3x}}}{3} \right)+x+C $ 

$=\dfrac{4}{3}{{e}^{3x}}+x+C$ 

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

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

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

$ =\dfrac{{{x}^{3}}}{3}-x+C$

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

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

$ =a\int{{{x}^{2}}dx+b\int{xdx+c\int{1.dx}}}$ 

$=a\left( \dfrac{{{x}^{3}}}{3} \right)+b\left( \dfrac{{{x}^{2}}}{2} \right)+cx+C $ 

9. $\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx}$

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

$ =2\int{{{x}^{2}}dx+\int{{{e}^{x}}dx}} $

$ =2\left( \dfrac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C$

$=\dfrac{2}{3}{{x}^{3}}+{{e}^{x}}+C $

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

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

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

$ =\int{xdx}+\int{\dfrac{1}{x}dx}-2\int{1.dx} $

$=\dfrac{{{x}^{2}}}{2}+\log \left| x \right|-2x+C $

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

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

\[ \int{\left( x+5-4{{x}^{-2}} \right)dx} \]

\[ =\int{xdx}+5\int{1.dx}-4\int{{{x}^{-2}}dx}\] 

\[=\dfrac{{{x}^{2}}}{2}+5x+\dfrac{4}{x}+C \]

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

Ans:$\int{\dfrac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

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

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

$ =\dfrac{2}{7}{{x}^{\dfrac{7}{2}}}+2{{x}^{\dfrac{3}{2}}}+8\sqrt{x}+C$

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

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

We obtain, on dividing:

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

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

$ =\dfrac{{{x}^{3}}}{3}+x+C$

14. $\int{\left( 1-x \right)}\sqrt{x}dx$

Ans: $\int{\left( 1-x \right)}\sqrt{x}dx$

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

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

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}-\dfrac{2}{5}{{x}^{\dfrac{5}{2}}}+C $

15. $\int{\sqrt{x}\left( 3{{x}^{2}}+2x+3 \right)dx}$

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

$ =3\int{\left( 2{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+3{{x}^{\dfrac{1}{2}}} \right)} $

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

$=\dfrac{6}{7}{{x}^{\dfrac{7}{2}}}+\dfrac{4}{5}{{x}^{\dfrac{5}{2}}}+2{{x}^{\dfrac{3}{2}}}+C $

16. $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

Ans: $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

\[ =2\int{xdx-3\int{\cos xdx}+\int{{{e}^{x}}dx}} \]

\[=\dfrac{2{{x}^{2}}}{2}-3\left( \sin x \right)+{{e}^{x}}+C \]

\[ ={{x}^{2}}-3\sin x+{{e}^{x}}+C\]

17. $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

Ans: $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

=\[ 2\int{{{x}^{2}}dx-3\int{\sin xdx+5\int{{{x}^{\dfrac{1}{2}}}}dx}}\]

= \[ \dfrac{2{{x}^{3}}}{3}-3\left( -\cos x \right)+5\left(\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C\] 

=$ \dfrac{2}{3}{{x}^{3}}+3\cos x+\dfrac{10}{3}{{x}^{\dfrac{3}{2}}}+C $

18. $\int{\sec x\left( \sec x+\tan x \right)dx}$

Ans: $\int{\sec x\left( \sec x+\tan x \right)dx}$

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

$ =\int{{{\sec }^{2}}xdx+\int{\sec x\tan xdx}}$

$ =\tan x+\sec x+C $

19. $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

Ans: $\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

$ =\int{\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}}dx} $

$=\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx}$

$=\int{{{\tan }^{2}}xdx}$

$ =\int{{{\sec }^{2}}xdx}-\int{1dx} $

$=\tan x-x+C $

20. $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

Ans: $\int{\dfrac{2-3\sin x}{{{\cos }^{2}}x}dx}$

$ =\int{\left( \dfrac{2}{{{\cos }^{2}}x}-\dfrac{3\sin x}{{{\cos }^{2}}x} \right)dx} $

$ =\int{2{{\sec }^{2}}xdx}-3\int{\tan x\sec xdx} $

$=2\tan x-3\sec x+C$

21. The anti – derivative of $\left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right)$ equals

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

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

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

4. $\dfrac{3}{2}{{x}^{\dfrac{3}{2}}}+\dfrac{1}{2}{{x}^{\dfrac{1}{2}}}+C$

Ans: 

 $ \left( \sqrt{x}+\dfrac{1}{\sqrt{x}} \right) $

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

$=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+2{{x}^{\dfrac{1}{2}}}+C $ 

 Thus, the correct answer is C.

22. If $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$ such that $f\left( 2 \right)=0$ then $f\left( x \right)$ is

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

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

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

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

Ans: Given, $\dfrac{d}{dx}f\left( x \right)=4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}$ 

Anti-derivative of $4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)$

\[ \therefore f\left( x \right)=\int{4{{x}^{3}}-\dfrac{3}{{{x}^{4}}}=f\left( x \right)} \]

$ f\left( x \right)=4\int{{{x}^{3}}dx-3\int{\left( {{x}^{-4}} \right)}dx} $

$f\left( x \right)=4\left( \dfrac{{{x}^{4}}}{4} \right)-3\left( \dfrac{{{x}^{-3}}}{-3} \right)+C $

$ f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}+C$

Also, 

$f\left( 2 \right)=0 $

$\therefore f\left( 2 \right)={{\left( 2 \right)}^{4}}+\dfrac{1}{{{\left( 2 \right)}^{3}}}+C=0 $ 

$ \Rightarrow 16+\dfrac{1}{8}+C=0 $ 

$ \Rightarrow C=\dfrac{-129}{8} $

$\therefore f\left( x \right)={{x}^{4}}+\dfrac{1}{{{x}^{3}}}-\dfrac{129}{8} $

 Thus, the correct answer is A.

Exercise 7.2

1. $\dfrac{2x}{1+{{x}^{2}}}$

Ans: Substitute $1+{{x}^{2}}=t$

$\therefore 2xdx=dt $

$\Rightarrow \int{\dfrac{2x}{1+{{x}^{2}}}dx}=\int{\dfrac{1}{t}}dt$

$ =\log \left| t \right|+C$

$ =\log \left| 1+{{x}^{2}} \right|+C$

$ =\log \left( 1+{{x}^{2}} \right)+C $

2. $\dfrac{{{\left( \log x \right)}^{2}}}{x}$

Ans: Substitute $\log \left| x \right|=t$

$ \therefore \dfrac{1}{x}dx=dt $

$\Rightarrow \int{\dfrac{{{\left( \log \left| x \right| \right)}^{2}}}{x}}dx=\int{{{t}^{2}}dt} $

$=\dfrac{{{t}^{3}}}{3}+C$

$ =\dfrac{{{\left( \log \left| x \right| \right)}^{3}}}{3}+C $

3. ${\dfrac{1}{x+x\log x}$

Ans: $\dfrac{1}{x+x\log x}=\dfrac{1}{x\left( 1+\log x \right)}$

Substitute $1+\log x=t$

$ \therefore \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x\left( 1+\log x \right)}dx}=\int{\dfrac{1}{t}dt} $

$ =\log \left| t \right|+C$

$=\log \left| 1+\log x \right|+C $          

4. ${Sinx.\sin \left( \cos x \right)$

Ans: $\operatorname{Sin}x.\sin \left( \cos x \right)$

 $ Put,\cos x=t $ 

 $ \therefore -\sin xdx=dt $ 

 $ \Rightarrow \int{\sin x.\sin \left( \cos x \right)dx}=-\int{\sin tdt} $

 $ =-\left[ -\cos t \right]+C $ 

 $ =\cos t+C $ 

 $ =\cos \left( \cos x \right)+C $

5. $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)$

Ans: $\operatorname{sin}\left( ax+b \right)\cos \left( ax+b \right)=\dfrac{2\sin \left( ax+b \right)\cos \left( ax+b \right)}{2}=\dfrac{\sin 2\left( ax+b \right)}{2}$ 

Substitute $2\left( ax+b \right)=t$

$ \therefore 2adx=dt $

$ \Rightarrow \int{\dfrac{\sin 2\left( ax+b \right)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin t}{2a}}dt $

$ =\dfrac{1}{4a}\left[ -\cos t \right]+C $

$ =\dfrac{-1}{4a}\cos 2\left( ax+b \right)+C$

6. $\sqrt{ax+b}$

Ans: Substitute $ax+b=t$

$ \Rightarrow adx=dt $

$ \therefore dx=\dfrac{1}{a}dt $

 $ \Rightarrow \int{{{\left( ax+b \right)}^{\dfrac{1}{2}}}dx}=\dfrac{1}{a}\int{{{t}^{\dfrac{1}{2}}}}dt $ 

 $=\dfrac{1}{a}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{3}{2}} \right)+C=\dfrac{2}{3a}{{\left( ax+b \right)}^{\dfrac{3}{2}}}+C $

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

Ans: Substitute $x+2=t$

$\therefore dx=dt $

$ \Rightarrow \int{x\sqrt{x+2}}=\int{\left( t-2 \right)\sqrt{t}}dt$

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

$ =\int{{{t}^{\dfrac{3}{2}}}dt-2\int{{{t}^{\dfrac{1}{2}}}}dt} $

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

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

Ans: Substitute $1+2{{x}^{2}}=t$

$ \therefore 4xdx=dt $

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

$ =\dfrac{1}{4}\int{{{t}^{\dfrac{1}{2}}}}dt $

$ =\dfrac{1}{4}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $

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

9. $\left( 4x+2 \right)\sqrt{{{x}^{2}}+x+1}$

Ans: Substitute ${{x}^{2}}+x+1=t$

$ \therefore \left( 2x+1 \right)dx=dt $

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

$ =\int{2\sqrt{t}dt}$

$=2\int{\sqrt{t}}dt $

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

10. $\dfrac{1}{x-\sqrt{x}}$

Ans: $\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)}$

Substitute $\left( \sqrt{x-1} \right)=t$

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

$ =2\log \left| t \right|+C$ 

$=2\log \left| \sqrt{x}-1 \right|+C$

11. $\dfrac{x}{\sqrt{x+4}},x>0$

Ans: Substitute $x+4=t$

$ \therefore dx=dt $

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

 $ =\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-4\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C=\dfrac{2}{3}{{\left( t \right)}^{\dfrac{3}{2}}}-8{{\left( t \right)}^{\dfrac{1}{2}}}+C $

$ =\dfrac{2}{3}t.{{t}^{\dfrac{1}{2}}}-8{{t}^{\dfrac{1}{2}}}+C $

 $ =\dfrac{2}{3}{{t}^{\dfrac{1}{2}}}\left( t-12 \right)+C $

$=\dfrac{2}{3}\sqrt{x+4}\left( x-8 \right)+C $

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

Ans: Substitute ${{x}^{3}}-1=t$

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

$ \Rightarrow \int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{5}}dx=\int{{{\left( {{x}^{3}}-1 \right)}^{\dfrac{1}{3}}}{{x}^{3}}.{{x}^{2}}dx}} $

$=\int{{{t}^{\dfrac{1}{3}}}\left( t+1 \right)\dfrac{dt}{3}=\dfrac{1}{3}\int{\left( {{t}^{\dfrac{4}{3}}}+{{t}^{\dfrac{1}{3}}} \right)dt}} $

$ =\dfrac{1}{3}\left[ \dfrac{3}{7}{{t}^{\dfrac{7}{3}}}+\dfrac{3}{4}{{t}^{\dfrac{4}{3}}} \right]+C $

$=\dfrac{1}{7}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{7}{3}}}+\dfrac{1}{4}{{\left( {{x}^{3}}-1 \right)}^{\dfrac{4}{3}}}+C $

13. $\dfrac{{{x}^{2}}}{{{\left( 2+3{{x}^{3}} \right)}^{3}}}$

Ans: Substitute $2+3{{x}^{3}}=t$

$ \therefore 9{{x}^{2}}dx=dt $

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

$ =\dfrac{1}{9}\left[ \dfrac{{{t}^{-2}}}{-2} \right]+C $

$=\dfrac{-1}{18{{\left( 2+3{{x}^{3}} \right)}^{2}}}+C $ 

14. $\dfrac{1}{x{{\left( \log x \right)}^{m}}},x>0$

Ans: Substitute $\log x=t$

$ \dfrac{1}{x}dx=dt$

$\Rightarrow \int{\dfrac{1}{x{{\left( \log x \right)}^{m}}}dx}=\int{\dfrac{dt}{{{\left( t \right)}^{m}}}=\left( \dfrac{{{t}^{-m-1}}}{1-m} \right)}+C $

$ =\dfrac{{{\left( \log x \right)}^{1-m}}}{\left( 1-m \right)}+C $

15. $\dfrac{x}{9-4{{x}^{2}}}$

Ans: Substitute $9-4{{x}^{2}}=t$

$ \therefore -8xdx=dt $

$ \Rightarrow \int{\dfrac{x}{9-4{{x}^{2}}}dx=\dfrac{-1}{8}\int{\dfrac{1}{t}dt}} $

$=\dfrac{-1}{8}\log \left| t \right|+C $

$=\dfrac{-1}{8}\log \left| 9-4{{x}^{2}} \right|+C$

16. ${{e}^{2x+3}}$

Ans: Substitute $2x+3=t$

$\therefore 2dx=dt $

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

$ =\dfrac{1}{2}\left( {{e}^{t}} \right)+C $

$ =\dfrac{1}{2}{{e}^{\left( 2x+3 \right)}}+C $

17. $\dfrac{x}{{{e}^{{{x}^{2}}}}}$

Ans: Substitute ${{x}^{2}}=t$

$ \therefore 2xdx=dt$

$ \Rightarrow \int{\dfrac{x}{{{e}^{{{x}^{2}}}}}dx=\dfrac{1}{2}\int{\dfrac{1}{{{e}^{t}}}dt}} $

$=\dfrac{1}{2}\int{{{e}^{-t}}dt} $

$ =\dfrac{1}{2}\left( \dfrac{{{e}^{-t}}}{-1} \right)+C$ 

$=-\dfrac{1}{2}{{e}^{-{{x}^{2}}}}+C $

$ =\dfrac{-1}{2{{e}^{{{x}^{2}}}}}+C $

18. $\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$

Ans: Substitute ${{\tan }^{-1}}x=t$

$ \therefore \dfrac{1}{1+{{x}^{2}}}dx=dt $ 

$ \Rightarrow \int{\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=dt} $ 

 $  ={{e}^{t}}+C $ 

$ ={{e}^{{{\tan }^{-1}}x}}+C$ 

19. Solve the following: $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Ans: Given expression $\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}$.

Let us substitute ${{e}^{2x}}+{{e}^{-2x}}=t$, we get

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

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

Integration of given expression is

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| t \right|+C$

Again substitute $t={{e}^{2x}}+{{e}^{-2x}}$, we get

$\therefore\int{\dfrac{{{e}^{2x}}-{{e}^{-2x}}}{{{e}^{2x}}+{{e}^{-2x}}}}=\dfrac{1}{2}\log \left| {{e}^{2x}}+{{e}^{-2x}} \right|+C$

20.  Solve the following: ${{\tan }^{2}}\left( 2x-3 \right)$.

Ans: Given expression ${{\tan }^{2}}\left( 2x-3 \right)$.

We can apply the identity ${{\tan }^{2}}x={{\sec }^{2}}x-1$, we get

${{\tan }^{2}}\left( 2x-3 \right)={{\sec }^{2}}\left( 2x-3 \right)-1$

Substitute $2x-3=t$, we get

$2dx=dt$ 

Integration of given expression is 

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\int{{{\sec }^{2}}\left( 2x-3 \right)-1dx}$

$\Rightarrow \int{{{\tan }^{2}}\left( 2x-3 \right)dx}=\dfrac{1}{2}\int{{{\sec }^{2}}tdt-\int{1dx}}$

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

Substitute $2x-3=t$

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

21.  Solve the following: ${{\sec }^{2}}\left( 7-4x \right)$.

Ans: Given expression ${{\sec }^{2}}\left( 7-4x \right)$.

Put $7-4x=t$, we get

$\therefore -4dx=dt$ 

Integration of given expression is

 $\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\int{{{\sec }^{2}}tdt}}$ 

$\Rightarrow \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan t+C}$

Substitute $7-4x=t$, we get

$\therefore \int{{{\sec }^{2}}\left( 7-4x \right)dx=-\dfrac{1}{4}\tan \left( 7-4x \right)+C}$

22.  Solve the following: $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.

Put ${{\sin }^{-1}}x=t$, we get

$\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx=\int{tdt}}$

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

Substitute ${{\sin }^{-1}}x=t$, we get

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

23.  Solve the following: $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Ans: Given expression is $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}$.

Given expression can be written as

 $\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}=\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}$

Let $3\cos x+2\sin x=t$, we get

$\left( -3\sin x+2\cos x \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{2\cos x-3\sin x}{2\left( 3\cos x+2\sin x \right)}}dx$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\int{\dfrac{dt}{2t}}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| t \right|+C$

Substitute $3\cos x+2\sin x=t$

$\therefore \int{\dfrac{2\cos x-3\sin x}{6\cos x+4\sin x}}dx=\dfrac{1}{2}\log \left| 2\sin x+3\cos x \right|+C$

24. Solve the following: $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Ans: Given expression $\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}$.

Given expression can be written as 

$\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}=\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}$

Let $\left( 1-\tan x \right)=t$, we get

$-{{\sec }^{2}}xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{{{\sec }^{2}}x}{{{\left( 1-\tan x \right)}^{2}}}}dx$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=\int{\dfrac{-dt}{{{t}^{2}}}}$

$\Rightarrow \int{\dfrac{1}{{{\cos }^{2}}x{{\left( 1-\tan x \right)}^{2}}}}dx=-\int{{{t}^{-2}}dt}$

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

Substitute $\left( 1-\tan x \right)=t$, 

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

25. Solve the following: $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Ans: Given expression is $\dfrac{\cos \sqrt{x}}{\sqrt{x}}$.

Let $\sqrt{x}=t$, we get

$\dfrac{1}{2\sqrt{x}}dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\int{\cos tdt}$

$\Rightarrow \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin t+C$

Substitute $\sqrt{x}=t$

$\therefore \int{\dfrac{\cos \sqrt{x}}{\sqrt{x}}dx}=2\sin \sqrt{x}+C$

26. Solve the following: $\sqrt{\sin 2x}\cos 2x$.

Ans: Given expression is $\sqrt{\sin 2x}\cos 2x$.

Let $\sin 2x=t$, we get

$2\cos 2xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\int{\sqrt{t}dt}}$ 

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{2}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C}$

$\Rightarrow \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+C}$

Substitute $\sin 2x=t$

$\therefore \int{\sqrt{\sin 2x}\cos 2xdx=\dfrac{1}{3}{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}+C}$

27. Solve the following: $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Ans: Given expression $\dfrac{\cos x}{\sqrt{1+\sin x}}$.

Let $1+\sin x=t$ 

$\therefore \cos xdx=dt$ 

Integration of given expression is  

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\int{\dfrac{dt}{\sqrt{t}}}$ 

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=\dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}+C$

$\Rightarrow \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{t}+C$

Substitute $1+\sin x=t$,

$\therefore \int{\dfrac{\cos x}{\sqrt{1+\sin x}}dx}=2\sqrt{1+\sin x}+C$

28.  Solve the following: $\cot x\log \sin x$.

Ans: Given expression $\cot x\log \sin x$.

Let $\log \sin x=t$, we get

$\dfrac{1}{\sin x}\cos xdx=dt$ 

$\Rightarrow \cot xdx=dt$ 

Integration of given expression is

$\int{\cot x\log \sin xdx=\int{tdt}}$ 

$\Rightarrow \int{\cot x\log \sin xdx=\dfrac{{{t}^{2}}}{2}+C}$

Substitute $\log \sin x=t$,

$\therefore \int{\cot x\log \sin xdx=\dfrac{1}{2}{{\left( \log \sin x \right)}^{2}}+C}$

29.  Solve the following: $\dfrac{\sin x}{1+\cos x}$.

Ans: Given expression $\dfrac{\sin x}{1+\cos x}$.

Let $1+\cos x=t$ 

$\therefore -\sin xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=\int{-\dfrac{dt}{t}}}$ 

$\Rightarrow \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| t \right|+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{1+\cos x}dx=-\log \left| 1+\cos x \right|+C}$

30.  Solve the following: $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Ans: Given expression $\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}$.

Let $1+\cos x=t$ 

$\therefore -\sin xdx=dt$ 

Integration of given expression is

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

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

$\Rightarrow \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{t}+C}$

Substitute $1+\cos x=t$,

$\therefore \int{\dfrac{\sin x}{{{\left( 1+\cos x \right)}^{2}}}dx=\dfrac{1}{1+\cos x}+C}$

.

31.Solve the following: $\dfrac{1}{1+\cot x}$.

Ans: Given expression $\dfrac{1}{1+\cot x}$.

Let  $I=\int{\dfrac{1}{1+\cot x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1+\cot x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1+\dfrac{\cos x}{\sin x}}}dx$

$\Rightarrow I=\int{\dfrac{\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\sin x}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)+\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \sin x+\cos x \right)}{\sin x+\cos x}+\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \sin x-\cos x \right)}{\sin x+\cos x}}}dx$

Let $\sin x+\cos x=t$ 

$\therefore \left( \cos x-\sin x \right)dx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\sin x+\cos x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \sin x+\cos x \right|+C$

32.  Solve the following: $\dfrac{1}{1-\tan x}$.

Ans: Given expression $\dfrac{1}{1-\tan x}$.

Let  $I=\int{\dfrac{1}{1-\tan x}}dx$

Integration of given expression is

$\Rightarrow I=\int{\dfrac{1}{1-\tan x}}dx$

$\Rightarrow I=\int{\dfrac{1}{1-\dfrac{\sin x}{\cos x}}}dx$

$\Rightarrow I=\int{\dfrac{\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{2\cos x}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)+\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( \cos x-\sin x \right)}{\cos x-\sin x}+\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}dx$

$\Rightarrow I=\dfrac{1}{2}\int{1dx+\dfrac{1}{2}\int{\dfrac{\left( \cos x+\sin x \right)}{\cos x-\sin x}}}dx$

Let $\cos x-\sin x=t$ 

$\therefore \left( -\sin x-\cos x \right)dx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\dfrac{1}{2}x+\dfrac{1}{2}\int{-\dfrac{dt}{t}}$

$\Rightarrow I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| t \right|+C$

Substitute $\cos x-\sin x=t$,

$\therefore I=\dfrac{x}{2}-\dfrac{1}{2}\log \left| \cos x-\sin x \right|+C$

33. Solve the following: $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Ans: Given expression $\dfrac{\sqrt{\tan x}}{\sin x\cos x}$.

Let $I=\int{\dfrac{\sqrt{\tan x}}{\sin x\cos x}dx}$ 

Multiply and divide by $\cos x$, we get

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\sin x\cos x\times \cos x}dx}$

$\Rightarrow I=\int{\dfrac{\sqrt{\tan x}\times \cos x}{\tan x\times {{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sec }^{2}}x}{\sqrt{\tan x}}dx}$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

Substitute in above obtained equation, we get

$\Rightarrow I=\int{\dfrac{dt}{\sqrt{t}}dx}$

$\Rightarrow I=2\sqrt{t}+C$

Substitute $\tan x=t$,

$\therefore I=2\sqrt{\tan x}+C$

34.  Solve the following: $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{{{\left( 1+\log x \right)}^{2}}}{x}$.

Let $1+\log x=t$ 

$\therefore \dfrac{1}{x}dx=dt$ 

Integration of given expression is

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

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

Substitute $1+\log x=t$

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

35. Solve the following: $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Ans: Given expression $\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}$.

Given expression can be written as

$\dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( \dfrac{x+1}{x} \right){{\left( x+\log x \right)}^{2}}$

$\Rightarrow \dfrac{\left( x+1 \right){{\left( x+\log x \right)}^{2}}}{x}=\left( 1+\dfrac{1}{x} \right){{\left( x+\log x \right)}^{2}}$

Let $x+\log x=t$ 

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

Integration of given expression is

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

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

Substitute $x+\log x=t$

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

36. Solve the following: $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$. 

Ans: Given expression $\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}$.

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

$\therefore 4{{x}^{3}}dx=dt$ 

Integration of given expression is

 $\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\dfrac{\sin \left( {{\tan }^{-1}}t \right)}{1+{{t}^{2}}}dt}}$ ………(1)

Let ${{\tan }^{-1}}t=u$ 

$\therefore \dfrac{1}{1+{{t}^{2}}}dt=du$ 

Substitute in eq. (1), we get

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\int{\sin udu}}$

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=\dfrac{1}{4}\left( -\cos u \right)+C}$

Substitute ${{\tan }^{-1}}t=u$,

$\Rightarrow \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}t \right)+C}$

Substitute ${{x}^{4}}=t$,

$\therefore \int{\dfrac{{{x}^{3}}\sin \left( {{\tan }^{-1}}{{x}^{4}} \right)}{1+{{x}^{8}}}dx=-\dfrac{1}{4}\cos \left( {{\tan }^{-1}}{{x}^{4}} \right)+C}$

37. $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}}dx$ equals 

1. ${{10}^{x}}-{{x}^{10}}+C$ 

2. ${{10}^{x}}+{{x}^{10}}+C$

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

4. $\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Ans: Given expression $\int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}$.

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

$\therefore \left( 10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10 \right)dx=dt$  

Integration of given expression is

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\int{\dfrac{dt}{t}}$ 

$\Rightarrow \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log t+C$

Substitute ${{x}^{10}}+{{10}^{x}}=t$,

$\therefore \int{\dfrac{10{{x}^{9}}+{{10}^{x}}{{\log }_{e}}10}{{{x}^{10}}+{{10}^{x}}}dx}=\log \left( {{10}^{x}}+{{x}^{10}} \right)+C$

Therefore, option D is the correct answer.

38. $\int{\dfrac{dx}{{{\sin }^{2}}{{\cos }^{2}}x}}$ equals

1. $\tan x+\cot x+C$ 

2. $\tan x-\cot x+C$

3. $\tan x\cot x+C$

4. $\tan x-\cot 2x+C$

Ans: Given expression $\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$.

Let  $I=\int{\dfrac{dx}{{{\sin }^{2}}x{{\cos }^{2}}x}}$ 

$I=\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}+\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow I=\int{\dfrac{1}{{{\cos }^{2}}x}dx}+\int{\dfrac{1}{{{\sin }^{2}}x}dx}$

$\Rightarrow I=\int{{{\sec }^{2}}xdx}+\int{cose{{c}^{2}}xdx}$

$\Rightarrow I=\tan x-\cot x+C$

Therefore, option B is the correct answer.

Exercise 7.3

1.  Solve the following: ${{\sin }^{2}}\left( 2x+5 \right)$.

Ans: Given expression ${{\sin }^{2}}\left( 2x+5 \right)$.

Given expression can be written as  

${{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos 2\left( 2x+5 \right)}{2}$

$\Rightarrow {{\sin }^{2}}\left( 2x+5 \right)=\dfrac{1-\cos \left( 4x+10 \right)}{2}$ 

Integration of given expression is

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\int{\dfrac{1-\cos \left( 4x+10 \right)}{2}dx}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}\int{1dx-\dfrac{1}{2}\int{\cos \left( 4x+10 \right)dx}}$

$\Rightarrow \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{2}\left( \dfrac{\sin \left( 4x+10 \right)}{4} \right)+C$

$\therefore \int{{{\sin }^{2}}\left( 2x+5 \right)dx}=\dfrac{1}{2}x-\dfrac{1}{8}\sin \left( 4x+10 \right)+C$

2. Solve the following: $\sin 3x\cos 4x$.

Ans: Given expression $\sin 3x\cos 4x$.

Using the identity $\sin A\cos B=\dfrac{1}{2}\left\{ \sin \left( A+B \right)+\sin \left( A-B \right) \right\}$ given expression can be written as

$\sin 3x\cos 4x=\dfrac{1}{2}\left\{ \sin \left( 3x+4x \right)+\sin \left( 3x-4x \right) \right\}$

Integration of above expression is

$\Rightarrow \int{\sin 3x\cos 4x}dx=\int{\dfrac{1}{2}\left\{ \sin 7x+\sin \left( -x \right) \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\left\{ \sin 7x+\sin x \right\}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\int{\sin 7xdx+\dfrac{1}{2}\int{\sin x}dx}$

$\Rightarrow \int{\sin 3x\cos 4x}dx=\dfrac{1}{2}\left( \dfrac{-\cos 7x}{7} \right)-\dfrac{1}{2}\left( -\cos x \right)+C$

$\therefore \int{\sin 3x\cos 4x}dx=\dfrac{-\cos 7x}{14}+\dfrac{\cos x}{2}+C$

3.  Solve the following: $\cos 2x\cos 4x\cos 6x$.

Ans: Given expression $\cos 2x\cos 4x\cos 6x$.

Using the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$ given expression can be written as

$\cos 2x\left( \cos 4x\cos 6x \right)=\cos 2x\dfrac{1}{2}\left\{ \cos \left( 4x+6x \right)+\cos \left( 4x-6x \right) \right\}$

Integration of the above expression is

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\cos 2x}\left[ \dfrac{1}{2}\left\{ \cos 10x+\cos \left( -2x \right) \right\} \right]dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\int{\left[ \dfrac{1}{2}\left\{ \cos 2x\cos 10x+\cos 2x\cos \left( -2x \right) \right\} \right]}dx$

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \cos 2x\cos 10x+{{\cos }^{2}}2x \right\} \right]}dx$

Again applying the identity $\cos A\cos B=\dfrac{1}{2}\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}$, we get

$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{2}\int{\left[ \left\{ \dfrac{1}{2}\cos \left( 2x+10x \right)+\cos \left( 2x-10x \right)+\left( \dfrac{1+\cos 4x}{2} \right) \right\} \right]}dx$$\Rightarrow \int{\cos 2x\left( \cos 4x\cos 6x \right)}=\dfrac{1}{4}\int{\left[ \cos 12x+\cos 8x+\cos 4x \right]}dx$

$\therefore \int{\cos 2x\cos 4x\cos 6x}=\dfrac{1}{4}\left[ \dfrac{\sin 12x}{12}+\dfrac{\sin 8x}{8}+\dfrac{\sin 4x}{4} \right]+C$

4. Solve the following: ${{\sin }^{3}}\left( 2x+1 \right)$.

Ans: Given expression ${{\sin }^{3}}\left( 2x+1 \right)$. 

Let $I=\int{{{\sin }^{3}}\left( 2x+1 \right)dx}$

$\Rightarrow I=\int{{{\sin }^{2}}\left( 2x+1 \right)\sin \left( 2x+1 \right)dx}$ 

$\Rightarrow I=\int{\left( 1-{{\cos }^{2}}\left( 2x+1 \right) \right)\sin \left( 2x+1 \right)dx}$

Let $\cos \left( 2x+1 \right)=t$ 

$\therefore -2\sin \left( 2x+1 \right)dx=dt$ 

Integration becomes

$\Rightarrow I=-\dfrac{1}{2}\int{\left( 1-{{t}^{2}} \right)dt}$

$\Rightarrow I=-\dfrac{1}{2}\left( t-\dfrac{{{t}^{3}}}{3} \right)+C$

Substitute $\cos \left( 2x+1 \right)=t$,

$\Rightarrow I=-\dfrac{1}{2}\left( \cos \left( 2x+1 \right)-\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3} \right)+C$

$\therefore \int{{{\sin }^{3}}\left( 2x+1 \right)dx}=\dfrac{-\cos \left( 2x+1 \right)}{2}+\dfrac{{{\cos }^{3}}\left( 2x+1 \right)}{3}+C$

5. Solve the following: ${{\sin }^{3}}x{{\cos }^{3}}x$.

Ans: Given expression ${{\sin }^{3}}x{{\cos }^{3}}x$.

Let $I=\int{{{\sin }^{3}}x{{\cos }^{3}}x}dx$ 

$\Rightarrow I=\int{{{\sin }^{2}}x\sin x{{\cos }^{3}}x}dx$

$\Rightarrow I=\int{{{\cos }^{3}}x\left( 1-{{\cos }^{2}}x \right)\sin x}dx$

Let $\cos x=t$ 

$\therefore -\sin dx=dt$ 

Integration becomes

$\Rightarrow I=-\int{{{t}^{3}}\left( 1-{{t}^{2}} \right)}dt$

$\Rightarrow I=-\int{\left( {{t}^{3}}-{{t}^{5}} \right)}dt$

$\Rightarrow I=-\left[ \dfrac{{{t}^{4}}}{4}-\dfrac{{{t}^{6}}}{6} \right]+C$

Substitute $\cos x=t$,

$\Rightarrow I=-\left[ \dfrac{{{\cos }^{4}}x}{4}-\dfrac{{{\cos }^{6}}x}{6} \right]+C$

$\therefore \int{{{\sin }^{3}}x{{\cos }^{3}}x}dx=\dfrac{{{\cos }^{6}}x}{6}-\dfrac{{{\cos }^{4}}x}{4}+C$

6. Solve the following: $\sin x\sin 2x\sin 3x$.  

Ans: Given expression $\sin x\sin 2x\sin 3x$.

Using the identity $\sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

$\Rightarrow \sin x\sin 2x\sin 3x=\sin x.\dfrac{1}{2}\cos \left( 2x-3x \right)-\cos \left( 2x+3x \right)$

Integration of given expression is

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos \left( -x \right)-\sin x\cos \left( 5x \right) \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\left( \sin x\cos x-\sin x\cos 5x \right)dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{2}\int{\dfrac{\sin 2x}{2}dx-\dfrac{1}{2}\int{\left( \sin x\cos 5x \right)}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{4}\left( \dfrac{-\cos 2x}{2} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{1}{2}\left( \sin \left( x+5x \right)+\sin \left( x-5x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin \left( -4x \right) \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{4}\int{\left\{ \left( \sin 6x+\sin 4x \right) \right\}dx}$

$\Rightarrow \int{\sin x\sin 2x\sin 3x}dx=\dfrac{-\cos 2x}{8}-\dfrac{1}{8}\left[ \dfrac{-\cos 6x}{3}+\dfrac{\cos 4x}{4} \right]+C$

$\therefore \int{\sin x\sin 2x\sin 3x}dx=\dfrac{1}{8}\left[ \dfrac{\cos 6x}{3}-\dfrac{\cos 4x}{4}-\cos 2x \right]+C$

7.  Solve the following: $\sin 4x\sin 8x$. 

Ans: Given expression $\sin 4x\sin 8x$.

Using the identity $\sin A\sin B=\dfrac{1}{2}\cos \left( A-B \right)-\cos \left( A+B \right)$, given expression can be written as

$\Rightarrow \sin 4x\sin 8x=\dfrac{1}{2}\cos \left( 4x-8x \right)-\cos \left( 4x+8x \right)$

Integration of given expression is

$\Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos \left( -4x \right)-\cos \left( 12x \right) \right)dx}$

$\Rightarrow \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx}$

$\therefore \int{\sin 4x\sin 8x}dx=\dfrac{1}{2}\left[ \dfrac{\sin 4x}{4}-\dfrac{\sin 12x}{12} \right]+C$

8.  Solve the following: $\dfrac{1-\cos x}{1+\cos x}$. 

Ans: Given expression $\dfrac{1-\cos x}{1+\cos x}$.

Using the identities $2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ given expression can be written as 

$\Rightarrow \dfrac{1-\cos x}{1+\cos x}=\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{1-\cos x}{1+\cos x}={{\tan }^{2}}\dfrac{x}{2}$

Integration of given expression is 

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\tan }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\int{\left[ {{\sec }^{2}}\dfrac{x}{2}-1 \right]dx}$

$\Rightarrow \int{\dfrac{1-\cos x}{1+\cos x}dx}=\left[ \dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}-x \right]+C$

$\therefore \int{\dfrac{1-\cos x}{1+\cos x}dx}=2\tan \dfrac{x}{2}-x+C$

9.  Solve the following: $\dfrac{\cos x}{1+\cos x}$. 

Ans: Given expression $\dfrac{\cos x}{1+\cos x}$.

Using the identity $\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ given expression can be written as 

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-\dfrac{{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}} \right]$

$\Rightarrow \dfrac{\cos x}{1+\cos x}=\dfrac{1}{2}\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right]$

Integration of given expression is 

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\tan }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 1-{{\sec }^{2}}\dfrac{x}{2}+1 \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\int{\left[ 2-{{\sec }^{2}}\dfrac{x}{2} \right]dx}$

$\Rightarrow \int{\dfrac{\cos x}{1+\cos x}dx}=\dfrac{1}{2}\left[ 2x-\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}} \right]+C$

$\therefore \int{\dfrac{\cos x}{1+\cos x}dx}=x-\tan \dfrac{x}{2}+C$

10.  Solve the following: ${{\sin }^{4}}x$. 

Ans: Given expression ${{\sin }^{4}}x$.

Given expression can be written as ${{\sin }^{4}}x={{\sin }^{2}}x{{\sin }^{2}}x$

$\Rightarrow {{\sin }^{4}}x=\left( \dfrac{1-\cos 2x}{2} \right)\left( \dfrac{1-\cos 2x}{2} \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}{{\left( 1-\cos 2x \right)}^{2}}$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 4x}{2}-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 4x}{2}-2\cos 2x \right)$

$\Rightarrow {{\sin }^{4}}x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right)$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\int{\left( \dfrac{3}{2}+\dfrac{1}{2}\cos 4x-2\cos 2x \right)}dx$

$\Rightarrow \int{{{\sin }^{4}}x}dx=\dfrac{1}{4}\left[ \dfrac{3}{2}x+\dfrac{\sin 4x}{8}-\sin 2x \right]+C$

$\therefore \int{{{\sin }^{4}}x}dx=\dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{1}{4}\sin 2x+C$

11.  Solve the following: ${{\cos }^{4}}2x$.

Ans: Given expression ${{\cos }^{4}}2x$.

Given expression can be written as  

${{\cos }^{4}}2x={{\left( {{\cos }^{2}}2x \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x={{\left( \dfrac{1+\cos 4x}{2} \right)}^{2}}$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+{{\cos }^{2}}4x+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1+\cos 8x}{2}+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right)$

$\Rightarrow {{\cos }^{4}}2x=\dfrac{1}{4}\left( \dfrac{3}{2}+\dfrac{\cos 8x}{2}+2\cos 4x \right)$

Integration of given expression is 

$\Rightarrow \int{{{\cos }^{4}}2xdx}=\int{\left( \dfrac{3}{8}+\dfrac{\cos 8x}{8}+\dfrac{\cos 4x}{2} \right)dx}$

$\therefore \int{{{\cos }^{4}}2xdx}=\dfrac{3}{8}x+\dfrac{\sin 8x}{64}+\dfrac{\sin 4x}{8}+C$

12. Solve the following: $\dfrac{{{\sin }^{2}}x}{1+\cos x}$.

Ans: Given expression $\dfrac{{{\sin }^{2}}x}{1+\cos x}$.

By applying the identity $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $\cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$, given expression can be written as 

  $\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{{{\left( 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=\dfrac{4{{\sin }^{2}}\dfrac{x}{2}{{\cos }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=2{{\sin }^{2}}\dfrac{x}{2}$

$\Rightarrow \dfrac{{{\sin }^{2}}x}{1+\cos x}=1-\cos x$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=\int{1dx-\int{\cos xdx}}$

$\therefore \int{\dfrac{{{\sin }^{2}}x}{1+\cos x}dx}=x-\sin x+C$

13.  Solve the following: $\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$. 

Ans: Given expression $\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$.

We can apply the identity $\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$ , we get

 $\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{-2\sin \dfrac{2x+2\alpha }{2}\sin \dfrac{2x-2\alpha }{2}}{-2\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \dfrac{2\left( x+\alpha  \right)}{2}\sin \dfrac{2\left( x-\alpha  \right)}{2}}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\sin \left( x+\alpha  \right)\sin \left( x-\alpha  \right)}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\dfrac{\left[ 2\sin \dfrac{x+\alpha }{2}\cos \dfrac{x+\alpha }{2} \right]\left[ 2\sin \dfrac{x-\alpha }{2}\cos \dfrac{x-\alpha }{2} \right]}{\sin \dfrac{x+\alpha }{2}\sin \dfrac{x-\alpha }{2}}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=4\cos \dfrac{x+\alpha }{2}\cos \dfrac{x-\alpha }{2}$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos \dfrac{x+\alpha }{2}+\dfrac{x-\alpha }{2}+\cos \dfrac{x+\alpha }{2}-\dfrac{x-\alpha }{2} \right]$

$\Rightarrow \dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=2\left[ \cos x+\cos \alpha  \right]$

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\int{\left[ \cos x+\cos \alpha  \right]}dx$

$\therefore \int{\dfrac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }dx}=2\left[ \sin x+x\cos \alpha  \right]+C$

14. Solve the following: $\dfrac{\cos x-\sin x}{1+\sin 2x}$. 

Ans: Given expression $\dfrac{\cos x-\sin x}{1+\sin 2x}$.

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Given expression can be written as  

$\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+\sin 2x}$ .

We can apply the identity $\sin 2x=2\sin x\cos x$, we get

 $\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}$

$\Rightarrow \dfrac{\cos x-\sin x}{1+\sin 2x}=\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}$

Let $\sin x+\cos x=t$ 

$\therefore \left( \cos x-\sin x \right)dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{\cos x-\sin x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}}$ 

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{\dfrac{dt}{{{t}^{2}}}}}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=\int{{{t}^{-2}}dt}}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-{{t}^{-1}}+C}$

$\Rightarrow \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{t}+C}$

Substitute $\sin x+\cos x=t$,

$\therefore \int{\dfrac{\cos x-\sin x}{1+\sin 2x}dx=-\dfrac{1}{\sin x+\cos x}+C}$

15.  Solve the following: ${{\tan }^{3}}2x\sec 2x$.

Ans: Given expression ${{\tan }^{3}}2x\sec 2x$.

Given expression can be written as

${{\tan }^{3}}2x\sec 2x={{\tan }^{2}}2x\tan 2x\sec 2x$

$\Rightarrow {{\tan }^{3}}2x\sec 2x=\left( {{\sec }^{2}}2x-1 \right)\tan 2x\sec 2x$

$\Rightarrow {{\tan }^{3}}2x\sec 2x={{\sec }^{2}}2x\tan 2x\sec 2x-\tan 2x\sec 2x$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\int{\tan 2x\sec 2xdx}$

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\int{{{\sec }^{2}}2x\tan 2x\sec 2xdx}-\dfrac{\sec 2x}{2}+C$

Let $\sec 2x=t$ 

$\therefore 2\sec 2x\tan 2xdx=dt$ 

Above integral becomes

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{1}{2}\int{{{t}^{2}}dt}-\dfrac{\sec 2x}{2}+C$

$\Rightarrow \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{t}^{3}}}{6}-\dfrac{\sec 2x}{2}+C$

Substitute $\sec 2x=t$,

$\therefore \int{{{\tan }^{3}}2x\sec 2xdx}=\dfrac{{{\left( \sec 2x \right)}^{3}}}{6}-\dfrac{\sec 2x}{2}+C$

16.  Solve the following: ${{\tan }^{4}}x$. 

Ans: Given expression ${{\tan }^{4}}x$.

Given expression can be written as 

$\Rightarrow {{\tan }^{4}}x={{\tan }^{2}}x{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x=\left( {{\sec }^{2}}x-1 \right){{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\tan }^{2}}x$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-\left( {{\sec }^{2}}x-1 \right)$

$\Rightarrow {{\tan }^{4}}x={{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1$

Integration of given expression is

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x-{{\sec }^{2}}x+1 \right)}dx$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{\left( {{\sec }^{2}}x{{\tan }^{2}}x \right)}dx-\int{{{\sec }^{2}}xdx+\int{1dx}}$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{\sec }^{2}}x{{\tan }^{2}}x}dx-\tan x+x+C$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

$\Rightarrow \int{{{\tan }^{4}}xdx}=\int{{{t}^{2}}}dt-\tan x+x+C$

$\Rightarrow \int{{{\tan }^{4}}xdx}=\dfrac{{{t}^{3}}}{3}-\tan x+x+C$

Substitute $\tan x=t$,

$\therefore \int{{{\tan }^{4}}xdx}=\dfrac{1}{3}{{\tan }^{3}}x-\tan x+x+C$

17. Solve the following: $\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

Ans: Given expression $\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$.

Given expression can be written as 

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{{{\sin }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}+\dfrac{{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}$

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\sin x}{{{\cos }^{2}}x}+\dfrac{\cos x}{{{\sin }^{2}}x}$

$\Rightarrow \dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}=\tan x\sec x+\cot xcosecx$

Integration of given expression is 

$\Rightarrow \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\tan x\sec xdx}+\int{\cot x\operatorname{cosecx}dx}$

$\therefore \int{\dfrac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\sec x-cosecx+C$

18.  Solve the following: $\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

Ans: Given expression $\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}$.

By applying the identity $\cos 2x=1-2{{\sin }^{2}}x$, we get

 $\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{\cos 2x+1-\cos 2x}{{{\cos }^{2}}x}$

$\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$

$\Rightarrow \dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}={{\sec }^{2}}x$

Integration of given expression is 

$\Rightarrow \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\int{{{\sec }^{2}}xdx}$

$\therefore \int{\dfrac{\cos 2x+2{{\sin }^{2}}x}{{{\cos }^{2}}x}}dx=\tan x+C$

19.  Solve the following: $\dfrac{1}{\sin x{{\cos }^{3}}x}$.

Ans: Given expression $\dfrac{1}{\sin x{{\cos }^{3}}x}$.

We can apply the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{{{\sin }^{2}}x}{\sin x{{\cos }^{3}}x}+\dfrac{{{\cos }^{2}}x}{\sin x{{\cos }^{3}}x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\dfrac{\sin x}{{{\cos }^{3}}x}+\dfrac{1}{\sin x\cos x}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\cos }^{2}}x}{\dfrac{\sin x\cos x}{{{\cos }^{2}}x}}$

$\Rightarrow \dfrac{1}{\sin x{{\cos }^{3}}x}=\tan x{{\sec }^{2}}x+\dfrac{{{\sec }^{2}}x}{\tan x}$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{\tan x{{\sec }^{2}}x}dx+\int{\dfrac{{{\sec }^{2}}x}{\tan x}}dx$

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\int{t}dt+\int{\dfrac{1}{\operatorname{t}}}dt$

$\Rightarrow \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{{{t}^{2}}}{2}+\log \left| t \right|+C$

Substitute $\tan x=t$,

$\therefore \int{\dfrac{1}{\sin x{{\cos }^{3}}x}}dx=\dfrac{1}{2}{{\tan }^{2}}x+\log \left| \tan x \right|+C$

20. Solve the following: $\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$. 

Ans: Given expression $\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}$.

Given expression can be written as 

$\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{{{\cos }^{2}}x+{{\sin }^{2}}x+2\sin x\cos x}$

We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $2\sin x\cos x=\sin 2x$, we get

 $\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+\sin 2x}$

Integration of given expression is

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

Let $1+\sin 2x=t$ 

$\therefore 2\cos 2xdx=dt$ 

Integration becomes

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$

$\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\log \left| t \right|+C$

Substitute $1+\sin 2x=t$

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

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

$\therefore \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\log \left| \left( \cos x+\sin x \right) \right|+C$

21.  Solve the following: ${{\sin }^{-1}}\left( \cos x \right)$. 

Ans: Given expression ${{\sin }^{-1}}\left( \cos x \right)$.

Let $\cos x=t$ 

$\therefore \sin x=\sqrt{1-{{t}^{2}}}$ 

$\Rightarrow -\sin xdx=dt$

$\Rightarrow dx=-\dfrac{dt}{\sin x}$

$\Rightarrow dx=-\dfrac{dt}{\sqrt{1-{{t}^{2}}}}$

Integration of given expression is

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

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

Let ${{\sin }^{-1}}t=u$ 

$\Rightarrow \dfrac{1}{\sqrt{1-{{t}^{2}}}}dt=du$ 

Integration becomes

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{4du}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{u}^{2}}}{2}+C$

Substitute ${{\sin }^{-1}}t=u$

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

Substitute $\cos x=t$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\left[ {{\sin }^{-1}}\left( \cos x \right) \right]}^{2}}}{2}+C$ ……..(1)

We know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$ 

$\therefore {{\sin }^{-1}}\left( \cos x \right)=\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \cos x \right)=\left( \dfrac{\pi }{2}-x \right)$ 

Substitute in eq. (1), we get

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{-{{\left( \dfrac{\pi }{2}-x \right)}^{2}}}{2}+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{1}{2}\left( \dfrac{{{\pi }^{2}}}{2}+{{x}^{2}}-\pi x \right)+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}\pi x+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+\left( C-\dfrac{{{\pi }^{2}}}{4} \right)$

$\therefore \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\dfrac{\pi x}{2}-\dfrac{{{x}^{2}}}{2}+{{C}_{1}}$

22. Solve the following: $\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$. 

Ans: Given expression $\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( x-b \right)\cos \left( x-a \right)-\cos \left( x-b \right)\sin \left( x-a \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]dx}$

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ -\log \left| \cos \left( x-b \right) \right|+\log \cos \left( x-a \right) \right]}$

$\Rightarrow \int{\dfrac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\dfrac{1}{\sin \left( a-b \right)}\left[ \log \left| \dfrac{\cos \left( x-a \right)}{\cos \left( x-b \right)} \right| \right]}+C$

23. Solve the following: $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$ is equal to

1. $\tan x+\cot x+C$ 

2. $\tan x+cosecx+C$ 

3. $-\tan x+\cot x+C$ 

4. $\tan x+\sec x+C$ 

Ans: Given expression $\int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.

Given expression can be written as

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\dfrac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}-\int{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{{{\sec }^{2}}xdx}-\int{cose{{c}^{2}}xdx}$

$\therefore \int{\dfrac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\tan x+\cot x+C$

Therefore, option A is the correct answer.

24. Solve the following: $\int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$ equals

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

2. $\tan \left( x{{e}^{x}} \right)+C$ 

3. $\tan \left( {{e}^{x}} \right)+C$

4. $\cot \left( {{e}^{x}} \right)+C$

Ans: Given expression $\int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$.

Let ${{e}^{x}}x=t$ 

$\therefore \left( {{e}^{x}}.x+{{e}^{x}}.1 \right)dx=dt$ 

$\Rightarrow {{e}^{x}}\left( x+1 \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{\dfrac{dt}{{{\cos }^{2}}t}}$

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{{{\sec }^{2}}t}dt$

$\Rightarrow \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan t+C$

Substitute ${{e}^{x}}x=t$,

$\therefore \int{\dfrac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan \left( {{e}^{x}}x \right)+C$

Therefore, option B is the correct answer.

Exercise 7.4

1. Solve the following: $\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}$.

Ans: Given expression $\dfrac{3{{x}^{2}}}{{{x}^{6}}+1}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

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

We know that $\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x}$ 

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

Substitute ${{x}^{3}}=t$,

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

2. Solve the following: $\dfrac{1}{\sqrt{1+4{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{1+4{{x}^{2}}}}$.

Let $2x=t$ 

$\therefore 2dx=dt$ 

Integration of given expression is

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

We know that $\int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{1+4{{x}^{2}}}}dx=\dfrac{1}{2}\log \left| t+\sqrt{{{t}^{2}}+1} \right|}+C$

Substitute $2x=t$,

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

3. Solve the following: $\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}$.

Ans: Given expression $\dfrac{1}{\sqrt{{{\left( 2-x \right)}^{2}}+1}}$.

Let $2-x=t$ 

$\therefore -dx=dt$ 

Integration of given expression is

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

We know that $\int{\dfrac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}dt=\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|$ 

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

Substitute $2-x=t$,

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

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

4. Solve the following: $\dfrac{1}{\sqrt{9-25{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{9-25{{x}^{2}}}}$.

Let $5x=t$ 

$\therefore 5dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{9-{{t}^{2}}}}dt}}$ 

$\Rightarrow \int{\dfrac{1}{\sqrt{9-25{{x}^{2}}}}dx=\dfrac{1}{5}\int{\dfrac{1}{\sqrt{{{3}^{2}}-{{t}^{2}}}}dt}}$

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

Substitute $5x=t$,

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

5.  Solve the following: $\dfrac{3x}{1+2{{x}^{4}}}$.

Ans: Given expression $\dfrac{3x}{1+2{{x}^{4}}}$.

Let $\sqrt{2}{{x}^{2}}=t$ 

$\therefore 2\sqrt{2}dx=dt$ 

Integration of given expression is

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

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

Substitute $\sqrt{2}{{x}^{2}}=t$,

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

6.  Solve the following: $\dfrac{{{x}^{2}}}{1-{{x}^{6}}}$.

Ans: Given expression $\dfrac{{{x}^{2}}}{1-{{x}^{6}}}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

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

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

Substitute ${{x}^{3}}=t$,

$\therefore \int{\dfrac{{{x}^{2}}}{1-{{x}^{6}}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}\log \left| \dfrac{1+{{x}^{3}}}{1-{{x}^{3}}} \right| \right]+C}$

7.  Solve the following: $\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}$.

Ans: Given expression $\dfrac{x-1}{\sqrt{{{x}^{2}}-1}}$.

Given expression can be written as

$\Rightarrow \dfrac{x-1}{\sqrt{{{x}^{2}}-1}}=\dfrac{x}{\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\sqrt{{{x}^{2}}-1}}$

Integration of given expression is

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

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

Let ${{x}^{2}}-1=t$ 

$\therefore 2xdx=dt$ 

Integration becomes

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

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

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

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

Substitute ${{x}^{2}}-1=t$

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

8.  Solve the following: $\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}$.

Ans: Given expression $\dfrac{{{x}^{2}}}{\sqrt{{{x}^{6}}+{{a}^{6}}}}$.

Let ${{x}^{3}}=t$ 

$\therefore 3{{x}^{2}}dx=dt$ 

Integration of given expression is

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

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

Substitute ${{x}^{3}}=t$,

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

9. Solve the following: $\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}$.

Ans: Given expression $\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}$.

Let $\tan x=t$ 

$\therefore {{\sec }^{2}}xdx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\int{\dfrac{dt}{\sqrt{{{t}^{2}}+{{2}^{2}}}}}}$ 

$\Rightarrow \int{\dfrac{{{\sec }^{2}}x}{\sqrt{{{\tan }^{2}}x+4}}dx=\log \left| t+\sqrt{{{\operatorname{t}}^{2}}+4} \right|}+C$

Substitute $\tan x=t$,

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

10.  Solve the following: $\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}$.

Ans: Given expression $\dfrac{1}{\sqrt{{{x}^{2}}+2x+2}}$.

Given expression can be written as 

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

Let $x+1=t$ 

$\therefore dx=dt$ 

Integration of given expression is

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

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

Substitute $x+1=t$,

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

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

11.  Solve the following: $\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}$.

Ans: Given expression $\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}$.

Given expression can be written as 

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

Let $3x+1=t$ 

$\therefore 3dx=dt$ 

Integration of given expression is

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

$\Rightarrow \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{t}{2} \right) \right]+C}$

Substitute $3x+1=t$,

$\therefore \int{\dfrac{1}{\sqrt{9{{x}^{2}}+6x+5}}dx=\dfrac{1}{3}\left[ \dfrac{1}{2}{{\tan }^{-1}}\left( \dfrac{3x+1}{2} \right) \right]+C}$

12.  Solve the following: $\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}$.

Given expression can be written as

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

$\Rightarrow \dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}=\dfrac{1}{\sqrt{16-\left( {{x}^{2}}+6x+9 \right)}}$

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

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

Let $x+3=t$ 

$\therefore dx=dt$ 

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{7-6x-{{x}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( t \right)}^{2}}}}dt}$

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

Substitute $x+3=t$,

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

13.  Solve the following: $\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}$.

Ans: Given expression $\dfrac{1}{\sqrt{\left( x-1 \right)\left( x-2 \right)}}$.

Given expression can be written as 

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

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

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

Let $x-\dfrac{3}{2}=t$ 

$\therefore dx=dt$ 

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

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

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

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

14.  Solve the following: $\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}$.

Ans: Given expression $\dfrac{1}{\sqrt{8+3x-{{x}^{2}}}}$.

Given expression can be written as

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

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

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

Let $x-\dfrac{3}{2}=t$ 

$\therefore dx=dt$ 

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

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

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

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

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

15.  Solve the following: $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$.

Ans: Given expression $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$.

Given expression can be written as

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{x}^{2}}-\left( a+b \right)x+\dfrac{{{\left( a+b \right)}^{2}}}{4}-\dfrac{{{\left( a+b \right)}^{2}}}{4}+ab}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-\dfrac{{{\left( a+b \right)}^{2}}}{4}}}$

$\Rightarrow \dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}=\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}$

Integration of given expression is

$\Rightarrow \int{\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}dx}=\int{\dfrac{1}{\sqrt{{{\left[ x-\dfrac{\left( a+b \right)}{4} \right]}^{2}}-{{\left( \dfrac{a+b}{2} \right)}^{2}}}}dx}$

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

$\therefore dx=dt$ 

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

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

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

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

16. $\dfrac{1}{\sqrt{\left( x-a \right)\left( x-b \right)}}$

Ans: $(x-a)(x-b)={{x}^{2}}-(a+b)x+ab$

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

Simplifying, 

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

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

Consider $\text{ }x-\left( \dfrac{a+b}{2} \right)=t\therefore dx=dt$

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

Using the logarithm formula of integration, 

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

Substitute the value of t, 

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

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

Ans:Consider $4x+1=A\dfrac{d}{dx}\left( 2{{x}^{2}}+x-3 \right)+B$

Simplifying, 

$\Rightarrow 4x+1=A(4x+1)+B$

$\Rightarrow 4x+1=4Ax+A+B$

We obtain the below values by equating the coefficients of $\text{x}$ and the constant term on both sides. 

$4~\text{A}=4\Rightarrow \text{A}=1$

$\text{A}+\text{B}=1\Rightarrow \text{B}=0$

Consider $2{{x}^{2}}+x-3=t$

$\therefore (4x+1)dx=dt$

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

Using the power rule of integration,

$=2\sqrt{t}+C$

Substitute the value of t,

$=2\sqrt{2{{x}^{2}}+x-3}+C$

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

 Ans:Consider  $\text{x}+2=A\dfrac{d}{dx}\left( {{x}^{2}}-1 \right)+B$

$\Rightarrow x+2=A(2x)+B......\left( 1 \right)$

We obtain the below values by equating the coefficients of x  and    the constant term on both sides. 

$2A=1\Rightarrow A=\dfrac{1}{2}$

$B=2$

From (1), we get

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

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

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

$\text{In }\dfrac{1}{2}\int{\dfrac{2x}{\sqrt{{{x}^{2}}-1}}}dx\text{ let }{{x}^{2}}-1=t\Rightarrow 2xdx=dt$

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

Integrating using the power rule

$=\dfrac{1}{2}[2\sqrt{t}]$

                   Simplifying,

$=\sqrt{t}\text{ }$

Substitute the value of t,

$=\sqrt{{{x}^{2}}-1}$

Then, $\int{\dfrac{2}{\sqrt{{{x}^{2}}-1}}}dx=2\int{\dfrac{1}{\sqrt{{{x}^{2}}-1}}}dx=2\log \left| x+\sqrt{{{x}^{2}}-1} \right|$

From equation (2), we get

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

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

Ans: Let $5x-2=A\dfrac{d}{dx}\left( 1+2x+3{{x}^{2}} \right)+B\text{ }$

$\Rightarrow \text{ }5\text{ }x-2=A\left( 2+6\text{ }x \right)+B......\left( 1 \right)$

We obtain the below values by equating the coefficients of x  and            

the constant term on both sides. 

$5=6A\Rightarrow A=\dfrac{5}{6}$

$2A+B=-2\Rightarrow B=-\dfrac{11}{3}$

Substitute the above values in (1)

$\therefore 5x-2=\dfrac{5}{6}(2+6x)+\left( -\dfrac{11}{3} \right)$

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

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

Consider ${{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx$

$\therefore\int{\dfrac{5x-2}{1+2x+3{{x}^{2}}}}dx=\dfrac{5}{6}{{I}_{1}}-\dfrac{11}{3}{{I}_{2}}...\left( 1 \right)$

${{I}_{1}}=\int{\dfrac{2+6x}{1+2x+3{{x}^{2}}}}dx$

Put $1+2x+3{{x}^{2}}=t$

$\Rightarrow (2+6x)dx=dt$

$\therefore {{I}_{1}}=\int{\dfrac{dt}{t}}$

Using the logarithm formula of integration,

${{I}_{1}}=\log |t|\text{ }$

Substitute the value of t,

${{I}_{1}}=\log \left| 1+2x+3{{x}^{2}} \right|...\left( 2 \right)$

Then, 

${{I}_{2}}=\int{\dfrac{1}{1+2x+3{{x}^{2}}}}dx$

$1+2x+3{{x}^{2}}$ can be rewritten as $1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right)$

Thus,

$1+3\left( {{x}^{2}}+\dfrac{2}{3}x \right)$

By completing square method, 

$=1+3\left( {{x}^{2}}+\dfrac{2}{3}x+\dfrac{1}{9}-\dfrac{1}{9} \right)$

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

Simplifying,

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

$=3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+\dfrac{2}{9} \right]$

$=3\left[ {{\left( x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right]$

Therefore ${{I}_{2}}$can be rewritten as , 

${{I}_{2}}=\dfrac{1}{3}\int{\dfrac{1}{\left. {{\left[ x+\dfrac{1}{3} \right)}^{2}}+{{\left( \dfrac{\sqrt{2}}{3} \right)}^{2}} \right]}}dx$

$=\dfrac{1}{3}\left[ \dfrac{3}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right) \right]$

Simplifying,

$=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \dfrac{3x+1}{\sqrt{2}} \right)...\left( 3 \right)$

We obtain the below values by substituting equations (2) and (3) in equation (1)

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

Simplifying,

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

20. $\dfrac{6x+7}{\sqrt{\left( x-5 \right)\left( x-4 \right)}}$

 Ans: Consider $6x+7=A\dfrac{d}{dx}\left( {{x}^{2}}-9x+20 \right)+B$

Differentiating, 

$\Rightarrow 6\text{ }x+7=A\left( 2\text{ }x-9 \right)+B$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$2~\text{A}=6\Rightarrow \text{A}=3$

$-9~\text{A}+\text{B}=7\Rightarrow \text{B}=34$

$\therefore 6\text{x}+7=3(2\text{x}-9)+34$

$\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=\int{\dfrac{3(2x-9)+34}{\sqrt{{{x}^{2}}-9x+20}}}dx$

$=3\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx+34\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Consider  ${{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

$\therefore\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}=3{{I}_{1}}+34{{I}_{2}}\,\,\,\,\,\,\,\,....(1)$

${{I}_{1}}=\int{\dfrac{2x-9}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Put  ${{x}^{2}}-9x+20=t$

$\Rightarrow (2x-9)dx=dt$

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}$

Integrating using the power rule

${{I}_{1}}=2\sqrt{t}$

Substitute the value of t,

${{I}_{1}}=2\sqrt{{{x}^{2}}-9x+20}\,\,\,.....(2)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}-9x+20}}}dx$

Consider 

${{x}^{2}}-9x+20$

By completing square methods, 

$={{x}^{2}}-9x+20+\dfrac{81}{4}-\dfrac{81}{4}$

$={{\left( x-\dfrac{9}{2} \right)}^{2}}-\dfrac{1}{4}$

$={{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}$

$\Rightarrow {{I}_{2}}=\int{\dfrac{1}{{{\left( x-\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}}}\,\,dx$

${{I}_{2}}=\log \left| \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right|.....\left( 3 \right)$

We obtain the below values by substituting equations (2) and (3) in (1), $\int{\dfrac{6x+7}{\sqrt{{{x}^{2}}-9x+20}}}dx=3\left[ 2\sqrt{{{x}^{2}}-9x+20} \right]+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+\text{C}$

Simplifying,

$=6\sqrt{{{x}^{2}}-9x+20}+34\log \left[ \left( x-\dfrac{9}{2} \right)+\sqrt{{{x}^{2}}-9x+20} \right]+C$

21. $\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}$

Ans:Consider, $\text{x}+2=A\dfrac{d}{dx}\left( 4x-{{x}^{2}} \right)+B$

$\Rightarrow x+2=A(4-2x)+B$

 We obtain the below values by equating the coefficients of x  and the  

constant term on both sides.

$-2A=1\Rightarrow A=-\dfrac{1}{2}$

 $4A+B=2\Rightarrow B=4$

$\Rightarrow (x+2)=-\dfrac{1}{2}(4-2x)+4$

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

Let ${{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx$ and ${{I}_{2}}\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx$

$\therefore \int{\dfrac{x+2}{\sqrt{4x-{{x}^{2}}}}}dx=-\dfrac{1}{2}{{I}_{1}}\text{ and }+4{{I}_{2}}\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{4-2x}{\sqrt{4x-{{x}^{2}}}}}dx$

 Let $4x-{{x}^{2}}=t$

$\Rightarrow (4-2x)dx=dt$

$\Rightarrow{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{4x-{{x}^{2}}}...\left( 2 \right)$

(Using the logarithm formula of integration,)

${{I}_{2}}=\int{\dfrac{1}{\sqrt{4x-{{x}^{2}}}}}dx$

Integrating using the power rule, 

$\Rightarrow 4x-{{x}^{2}}=-\left( -4x+{{x}^{2}} \right)$

By completing square methods, 

$=\left( -4x+{{x}^{2}}+4-4 \right)$

$=4-{{(x-2)}^{2}}$

$={{(2)}^{2}}-{{(x-2)}^{2}}$

$\therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(2)}^{2}}-{{(x-2)}^{2}}}}}dx={{\sin }^{-1}}\left( \dfrac{x-2}{2} \right)...\left( 3 \right)$

   Using equations (2) and (3) in (1), to obtain 

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

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

22.$\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}$

Ans:$\int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}\int{\dfrac{2(x+2)}{\sqrt{{{x}^{2}}+2x+3}}}dx$

Simplifying,

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

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

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

Let ${{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx\,\,\,and\,\,{{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx$

$\therefore \int{\dfrac{x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx=\dfrac{1}{2}{{I}_{1}}+{{I}_{2}}\,\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x+2}{\sqrt{{{x}^{2}}+2x+3}}}dx$

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

Integrating using the power rule, 

$\Rightarrow (2\text{x}+2)\text{dx}=\text{dt}\,\,\,\,\,\,$

$\,{{I}_{1}}=\int{\dfrac{dt}{\sqrt{t}}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+2x+3}..\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+2x+3}}}dx.$

By completing square methods,

$\Rightarrow {{x}^{2}}+2x+3={{x}^{2}}+2x+1+2={{(x+1)}^{2}}+{{(\sqrt{2})}^{2}}$

$\therefore {{I}_{2}}=\int{\dfrac{1}{\sqrt{{{(x+1)}^{2}}+{{(\sqrt{2})}^{2}}}}}dx=\log \left| (x+1)+\sqrt{{{x}^{2}}+2x+3} \right|...\left( 3 \right)$

Using equations (2) and (3) in (1), to obtain                             

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

23. $\dfrac{x+3}{{{x}^{2}}-2x-5}$

Ans:  Consider $(x+3)=A\dfrac{d}{dx}\left( {{x}^{2}}-2x-5 \right)+B$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

We obtain the below values by equating the coefficients of x  and the constant term on both sides.

$2A=1\Rightarrow A=\dfrac{1}{2}$

$-2A+B=3\Rightarrow B=4$

$\therefore (x+3)=\dfrac{1}{2}(2x-2)+4$

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

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

Consider${{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx$

$\therefore \int{\dfrac{x+3}{{{x}^{2}}-2x-5}}dx=\dfrac{1}{2}{{I}_{1}}+4{{I}_{2}}...\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x-2}{{{x}^{2}}-2x-5}}dx$

Put ${{x}^{2}}-2x-5=\text{t}$

$\Rightarrow (2x-2)dx=dt$

Using the logarithm formula of integration,

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}-2x-5 \right|\,\,\,....\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{{{x}^{2}}-2x-5}}dx$

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

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

$=\dfrac{1}{2\sqrt{6}}\log \left( \dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}} \right)....\left( 3 \right)$

We obtain the below values by substituting (2) and (3) in (1), 

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

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

24. $\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}$

Ans:  $5x+3=A\dfrac{a}{dx}\left( {{x}^{2}}+4x+10 \right)+B$

$\Rightarrow 5x+3=A(2x+4)+B$

Equating the coefficients of $\text{x}$and constant term, we get

$2A=5\Rightarrow A=\dfrac{5}{2}$

$4A+B=3\Rightarrow B=-7$

$\therefore 5x+3=\dfrac{5}{2}(2x+4)-7$

$\Rightarrow \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\int{\dfrac{\dfrac{5}{2}(2x+4)-7}{\sqrt{{{x}^{2}}+4x+10}}}dx$

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

Let ${{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx$ and ${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx$

$\therefore \int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}{{I}_{1}}-7{{I}_{2}}\,\,\,\,\,\,\,\,\,....\left( 1 \right)$

Then, ${{I}_{1}}=\int{\dfrac{2x+4}{\sqrt{{{x}^{2}}+4x+10}}}dx$

Consider ${{x}^{2}}+4x+10=\text{t}\therefore (2x+4)dx=dt$

$\Rightarrow {{I}_{1}}=\int{\dfrac{dt}{t}}=2\sqrt{t}=2\sqrt{{{x}^{2}}+4x+10}\,\,\,\,\,\,\,......\left( 2 \right)$

${{I}_{2}}=\int{\dfrac{1}{\sqrt{{{x}^{2}}+4x+10}}}dx$

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

$=\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|...\left( 3 \right)$

We obtain the below values by using equations (2) and (3) in (1). $\int{\dfrac{5x+3}{\sqrt{{{x}^{2}}+4x+10}}}dx=\dfrac{5}{2}\left[ 2\sqrt{{{x}^{2}}+4x+10} \right]-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$$=5\sqrt{{{x}^{2}}+4x+10}-7\log \left| (x+2)\sqrt{{{x}^{2}}+4x+10} \right|+C$

25. $\int{\dfrac{dx}{{{x}^{2}}+2x+2}}$equals 

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

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

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

4. ${{\tan }^{-1}}x+C$

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

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

$=\left[ {{\tan }^{-1}}(x+1) \right]+C$

Hence, the right response is is B.

26. $\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$ equals 

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

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

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

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

Ans:

             $\int{\dfrac{dx}{\sqrt{9x-4{{x}^{2}}}}}$

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

    By completing square methods,

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

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

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

$=\dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{x-\dfrac{9}{8}}{\dfrac{9}{8}} \right) \right]+C\quad \left( \int{\dfrac{dy}{\sqrt{{{a}^{2}}-{{y}^{2}}}}}={{\sin }^{-1}}\dfrac{y}{a}+C \right)$

   Simplifying,

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

Hence, the right response is B.

Exercise 7.5:

1. $\dfrac{x}{\left( x+1 \right)\left( x+2 \right)}$

Ans:   Let $\dfrac{x}{(x+1)(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$\Rightarrow x=A(x+2)+B(x+1)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$\text{A}+\text{B}=1$

$2~\text{A}+\text{B}=0$

On solving, we get

$\text{A}=-1\text{ and B}=2$

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

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

Using the logarithm formula of integration,

$=-\log |x+1|+2\log |x+2|+C$

$=\log {{(x+2)}^{2}}-\log |x+1|+C$

Simplifying,

$=\log \dfrac{{{(x+2)}^{2}}}{(x+1)}+C$

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

Ans:  Let $\dfrac{1}{(x+3)(x-3)}=\dfrac{A}{(x+3)}+\dfrac{B}{(x-3)}$

$1=A\left( x-3 \right)+B\left( x+3 \right)$

Equating the coefficients of $\text{x}$and constant term, we get

$A+B=0$ 

$1=-3A+3B$

$A=-\dfrac{1}{6}\text{ and }B=\dfrac{1}{6}$

$\therefore \dfrac{1}{(x+3)(x-3)}=\dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)}$

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

Using the logarithm formula of integration,

$=-\dfrac{1}{6}\log |x+3|+\dfrac{1}{6}\log |x-3|+C=\dfrac{1}{6}\log \dfrac{|(x-3)|}{|(x+3)|}+C$

3. $\dfrac{3x-1}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans: Let $\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\quad \ldots \left( 1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$\text{A}+\text{B}+\text{C}=0$

$-5~\text{A}-4~\text{B}-3\text{C}=3$

$6~\text{A}+3~\text{B}+2\text{C}=-1$

Solving these equations, to obtain 

$\text{A}=1,~\text{B}=-5,\text{ and C}=4$

$\therefore \dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)}$

$\Rightarrow \int{\dfrac{3x-1}{(x-1)(x-2)(x-3)}}dx=\int{\left\{ \dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)} \right\}}dx$

Using the logarithm formula of integration,

$=\log |x-1|-5\log |x-2|+4\log |x-3|+C$

4. $\dfrac{x}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}$

Ans:Let $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$x=A\left( x-2 \right)\left( x-3 \right)+B\left( x-1 \right)\left( x-3 \right)+C\left( x-1 \right)\left( x-2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0\text{ }$

$-5\text{ }A-4\text{ }B-3\text{ }C=1$

$6\text{ }A+4\text{ }B+2\text{ }C=0$

Solving these equations, to obtain 

$A=\dfrac{1}{2},B=2\text{ and }C=\dfrac{3}{2}$

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

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

Using the logarithm formula of integration,

$=\dfrac{1}{2}\log |x-1|-2\log |x-2|+\dfrac{3}{2}\log |x-3|+C$

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

 Ans:$\dfrac{2x}{{{x}^{2}}+3x+2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$2\text{ }x=A\left( x+2 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B=2$

$2\text{ }A+B=0$

Solving these equations, we get

$A=-2\text{ and }B=4$

$\therefore \dfrac{2x}{(x+1)(x+2)}=\dfrac{-2}{(x+1)}+\dfrac{4}{(x+2)}$

$\Rightarrow \int{\dfrac{2x}{(x+1)(x+2)}}dx=\int{\left\{ \dfrac{4}{(x+2)}-\dfrac{2}{(x+1)} \right\}}dx$

Using the logarithm formula of integration,

$=4\log |x+2|-2\log |x+1|+C$

6. $\dfrac{1-{{x}^{2}}}{x\left( 1-2x \right)}$

Ans:  It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $\left( 1-{{x}^{2}} \right)$ by $x(1-2x)$ to obtain, $\dfrac{1-{{x}^{2}}}{x(1+2x)}=\dfrac{1}{2}+\dfrac{1}{2}\left( \dfrac{2-x}{x(1-2x)} \right)$

Let $\dfrac{2-x}{x(1-2x)}=\dfrac{A}{x}+\dfrac{B}{(1-2x)}...\left( 1 \right)$

$\Rightarrow (2-x)=A(1-2x)+Bx$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

 $-2~\text{A}+\text{B}=-1$ and, $A=2$

Solving these equations, to obtain $A=2\text{ and }B=3$

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

Substituting in equation (1), we get

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

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

Using the power rule and logarithm formula of integration,

$=\dfrac{x}{2}+\log |x|+\dfrac{3}{2(-2)}\log |1-2x|+C=\dfrac{x}{2}+\log |x|-\dfrac{3}{4}\log |1-2x|+C$

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

Ans: Let $\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{Ax+B}{\left( {{x}^{2}}+1 \right)}+\dfrac{C}{(x-1)}...\left( 1 \right)$

$x=(Ax+B)(x-1)+C\left( {{x}^{2}}+1 \right)$

$x=A{{x}^{2}}-Ax+Bx-B+C{{x}^{2}}+C$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0\text{ }$

$-A+B=1$

$-B+C=0$

On solving these equations, to obtain  $A=-\dfrac{1}{2},B=\dfrac{1}{2},and\,\,C=\dfrac{1}{2}$

From equation (1), to obtain  

$\therefore \dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}=\dfrac{\left( -\dfrac{1}{2}x+\dfrac{1}{2} \right)}{{{x}^{2}}+1}+\dfrac{\dfrac{1}{2}}{(x-1)}$

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

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

Consider $\int{\dfrac{2x}{{{x}^{2}}+1}}dx,lel\left( {{x}^{2}}+1 \right)=t\Rightarrow 2xdx=dt$

$\Rightarrow \int{\dfrac{2x}{{{x}^{2}}+1}}dx=\int{\dfrac{dt}{t}}=\log |t|=\log \left| {{x}^{2}}+1 \right|$

$\therefore \int{\dfrac{x}{\left( {{x}^{2}}+1 \right)(x-1)}}=-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+\dfrac{1}{2}\log |x-1|+C$

$=\dfrac{1}{2}\log |x-1|-\dfrac{1}{4}\log \left| {{x}^{2}}+1 \right|+\dfrac{1}{2}{{\tan }^{-1}}x+C$

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

Ans: $\dfrac{x}{{{(x-1)}^{2}}(x+2)}$

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

$x=A(x-1)(x+2)+B(x+2)+C{{(x-1)}^{2}}$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+C=0$

$A+B-2\text{ }C=1$

On solving, to obtain

$A=\dfrac{2}{9}\text{ and }C=\dfrac{-2}{9}$

$B=\dfrac{1}{3}$

$\therefore \dfrac{x}{{{(x-1)}^{2}}(x+2)}=\dfrac{2}{9(x-1)}+\dfrac{1}{3{{(x-1)}^{2}}}-\dfrac{2}{9(x\mid 2)}$

$\Rightarrow \int{\dfrac{x}{{{(x-1)}^{2}}(x+2)}}dx=\dfrac{2}{9}\int{\dfrac{1}{(x-1)}}dx+\dfrac{1}{3}\int{\dfrac{1}{{{(x-1)}^{2}}}}dx-\dfrac{2}{9}\int{\dfrac{1}{(x-2)}}dx$

Using the power rule and logarithm formula of integration,

$=\dfrac{2}{9}\log |x-1|+\dfrac{1}{3}\left( \dfrac{-1}{x-1} \right)-\dfrac{2}{9}\log |x+2|+C$

Simplifying,

$=\dfrac{2}{9}\log \left| \dfrac{x-1}{x+2} \right|-\dfrac{1}{3(x-1)}+C$

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

Ans:   $\dfrac{3x+5}{{{x}^{3}}-{{x}^{2}}-x+1}=\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}$

let  $\dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{A}{(x+1)}+\dfrac{B}{{{(x-1)}^{2}}}+\dfrac{C}{(x+1)}$

$3x+5=A(x-1)(x+1)+B(x+1)+{{(x-1)}^{2}}$

$3x+5=A{{(x-1)}^{2}}+B(x+1)+C\left( {{x}^{2}}+1-2x \right)\quad \ldots (1)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and  

the constant term on both sides.

$A+C=0\text{ }$

$B-2\text{ }C=3\text{ }$

$-A+B+C=5$

On solving, to obtain   $B=4\,\,A=-\dfrac{1}{2}\,\,and\,\,C=\dfrac{1}{2}$

$\therefore \dfrac{3x+5}{{{(x-1)}^{2}}(x+1)}=\dfrac{-1}{2(x-1)}+\dfrac{4}{{{(x-1)}^{2}}}+\dfrac{1}{2(x+1)}$

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

Using the power rule and logarithm formula of integration,

$=-\dfrac{1}{2}\log |x-1|+4\left( \dfrac{-1}{x-1} \right)+\dfrac{1}{2}\log |x+1|+C$

$=\dfrac{1}{2}\log \left| \dfrac{x+1}{x-1} \right|-\dfrac{4}{(x-1)}+C$

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

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

Let $\dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{C}{(2x+3)}$

$\Rightarrow (2x-3)=A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)$

$\Rightarrow (2x-3)-A\left( 2{{x}^{2}}+x-3 \right)+B\left( 2{{x}^{2}}+5x-3 \right)+C\left( {{x}^{2}}-1 \right)$

$\Rightarrow (2x-3)=(2A+2B+C){{x}^{2}}+(A+5B)x+(-3A+3B-C)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$2~\text{A}+2~\text{B}+\text{C}=0$

$\text{A}+5~\text{B}=2$

$-3~\text{A}+3~\text{B}-\text{C}=-3$

On solving, to obtain $B=-\dfrac{1}{10},A=\dfrac{5}{2},\text{ and C}=-\dfrac{24}{5}$

$\therefore \dfrac{2x-3}{(x+1)(x-1)(2x+3)}=\dfrac{5}{2(x+1)}-\dfrac{1}{10(x-1)}-\dfrac{24}{5(2x+3)}$

$\Rightarrow \int{\dfrac{2x-3}{\left( {{x}^{2}}-1 \right)(2x+3)}}dx=\dfrac{5}{2}\int{\dfrac{1}{(x+1)}}dx-\dfrac{1}{10}\int{\dfrac{1}{x-1}}dx-\dfrac{24}{5}\int{\dfrac{1}{(2x+3)}}dx$

Using the logarithm formula of integration,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{24}{5\times 2}\log |2x+3|$

Simplifying,

$=\dfrac{5}{2}\log |x+1|-\dfrac{1}{10}\log |x-1|-\dfrac{12}{5}\log |2x+3|+C$

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

Ans: $\dfrac{5x}{(x+1)\left( {{x}^{2}}-4 \right)}=\dfrac{5x}{(x+1)(x+2)(x-2)}$

let $\dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-2)}$

$5\text{ }x=A\left( x+2 \right)\left( x-2 \right)+B\left( x+1 \right)\left( x-2 \right)+C\left( x+1 \right)\left( x+2 \right)$

We obtain the below values by equating the coefficients of x , ${{x}^{2}}$ and the constant term on both sides.

$A+B+C=0$

$-B+3C=5\text{ and, }-4A-2B+2C=0$

On solving, to obtain

$A=\dfrac{5}{3},B=-\dfrac{5}{2},\text{ and }C=\dfrac{5}{6}$

$\therefore \dfrac{5x}{(x+1)(x+2)(x-2)}=\dfrac{5}{3(x+1)}+-\dfrac{5}{2(x+2)}+\dfrac{5}{6(x-2)}$

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

Using the logarithm formula of integration,

$=\dfrac{5}{3}\log |x+1|-\dfrac{5}{2}\log |x+2|+\dfrac{5}{6}\log |x-2|+C$

12. $\dfrac{{{x}^{2}}+x+1}{{{x}^{2}}-1}$

 Ans:  On dividing $\left( {{x}^{3}}+x+1 \right)\,\,\,by\,\,{{x}^{2}}-1,$we get

$\dfrac{{{x}^{3}}+x+1}{{{x}^{2}}-1}=x+\dfrac{2x+1}{{{x}^{2}}-1}$

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

$2\text{ }x+1=A\left( x-1 \right)+B\left( x+1 \right)$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$A+B=2$

$-A+B=1$

On solving, to obtain  

$A=\dfrac{1}{2}\text{ and }B=\dfrac{3}{2}\text{ }$

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

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

$=\dfrac{{{x}^{2}}}{2}+\log |x+1|+\dfrac{3}{2}\log |x-1|+C$

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

Ans: Let $\dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(1-x)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}$

$2=A\left( 1+{{x}^{2}} \right)+(Bx+C)(1-x)$

$2=A+A{{x}^{2}}+Bx-B{{x}^{2}}+C-Cx$

We obtain the below values by equating the coefficients of x ${{x}^{2}}$ and the constant term on both sides. 

$\text{A}-\text{B}=0$

$B-C=0$

$A+C=2$

On solving these equations, to obtain

$A=1,B=1\text{, and }C=1$

$\therefore \dfrac{2}{(1-x)\left( 1+{{x}^{2}} \right)}=\dfrac{1}{1-x}+\dfrac{x+1}{1+{{x}^{2}}}$

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

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

$=-\log |x-1|+\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+{{\tan }^{-1}}x+C$

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

Ans:   Let $\dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{A}{(x+2)}+\dfrac{B}{{{(x+2)}^{2}}}$

$\Rightarrow 3x-1=A(x+2)+B$

We obtain the below values by equating the coefficients of x  and the constant term on both sides. 

$A=3$

$2A+B=-1\Rightarrow B=-7$

$\therefore \dfrac{3x-1}{{{(x+2)}^{2}}}=\dfrac{3}{(x+2)}-\dfrac{7}{{{(x+2)}^{2}}}$

$\Rightarrow \int{\dfrac{3x-1}{{{(x+2)}^{2}}}}dx=3\int{\dfrac{1}{(x+2)}}dx-7\int{\dfrac{x}{{{(x+2)}^{2}}}}dx$

Using the power rule and logarithm formula of integration

$=3\log |x+2|-7\left( \dfrac{-1}{(x+2)} \right)+C$

$=3\log |x+2|+\dfrac{7}{(x+2)}+C$

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

Ans:  $\dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{1}{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)}=\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}$

Let $\dfrac{1}{(x+1)(x-1)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{Cx+D}{\left( {{x}^{2}}+1 \right)}$ 

$1=A(x-1)\left( 1+{{x}^{2}} \right)+B(x+1)\left( 1+{{x}^{2}} \right)+(Cx+D)\left( {{x}^{2}}-1 \right)$

$1=A\left( {{x}^{3}}+x-{{x}^{2}}-1 \right)+B\left( {{x}^{3}}+x+{{x}^{2}}+1 \right)+C{{x}^{3}}+D{{x}^{2}}-Cx-D$

$1=(A+B+C){{x}^{3}}+(-A+B+D){{x}^{2}}+(A+B-C)x+(-A+B-D)$

We obtain the below values by equating the coefficients of${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x},$and constant term, we get

$A+B+C=0\text{ }$

$-A+B+D=0$

$A+B-C=0\text{ }$

$-A+B-D=1$

$A=-\dfrac{1}{4},B=\dfrac{1}{4},C=0,\text{ and }D=-\dfrac{1}{2}$

$\therefore \dfrac{1}{\left( {{x}^{4}}-1 \right)}=\dfrac{-1}{4(x+1)}+\dfrac{1}{4(x-1)}+\dfrac{1}{2\left( {{x}^{2}}+1 \right)}$

$\Rightarrow \int{\dfrac{1}{{{x}^{4}}-1}}dx=-\dfrac{1}{4}\log |x-1|+\dfrac{1}{4}\log |x-1|-\dfrac{1}{2}{{\tan }^{1}}x+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{x-1}{x+1} \right|-\dfrac{1}{2}{{\tan }^{1}}x+C$

16. $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$ $\text{ [ }$Hint: Multiply numerator and denominator by ${{x}^{n-1}}$ and put ${{x}^{n}}=t$ $\text{]}$

Ans: $\dfrac{1}{x\left( {{x}^{n}}+1 \right)}$

Numerator and denominator are multiplied by ${{x}^{n-1}}$, to obtain 

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

Consider  ${{x}^{n}}=t\Rightarrow {{x}^{n-1}}dx=dt$

$\therefore \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\int{\dfrac{{{x}^{n-1}}}{{{x}^{n}}\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\dfrac{1}{t(t+1)}}dt$

$\text{ Let }\dfrac{1}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{(t+1)}$

$1=A\left( 1+t \right)+B\text{ }t$

We obtain the below values by equating the coefficients of $\text{t}$and constant, 

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{t(t+1)}=\dfrac{1}{t}-\dfrac{1}{(1+t)}$

$\Rightarrow \int{\dfrac{1}{x\left( {{x}^{n}}+1 \right)}}dx=\dfrac{1}{n}\int{\left\{ \dfrac{1}{t}-\dfrac{1}{(1+t)} \right\}}dx$

$=\dfrac{1}{n}[\log |t|-\log |t+1|]+C$

Substitute the value of t,

$=-\dfrac{1}{n}\left[ \log \left| {{x}^{n}} \right|-\log \left| {{x}^{n}}+1 \right| \right]+C$

Simplifying,

$=\dfrac{1}{n}\log \left| \dfrac{{{x}^{u}}}{{{x}^{\prime \prime }}+1} \right|+C$

17. $\dfrac{\cos x}{\left( 1-\sin x \right)\left( 2-\sin x \right)}$ $\text{[}$ hint: Put $\sin x=t$ $\text{]}$

Ans:  $\dfrac{\cos x}{(1-\sin x)(2-\sin x)}\,\,\,\,Put,\sin x=t\Rightarrow \cos xdx=dt$

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

let $\dfrac{1}{(1-t)(2-t)}=\dfrac{A}{(1-t)}+\dfrac{B}{(2-t)}$ 

$1=A\left( 2-t \right)+B\left( 1-t \right)$

We obtain the below values by equating the coefficients of t and constant, 

$-2~\text{A}-\text{B}=0\,\,,and\,\,2~\text{A}+\text{B}=1$

$A=1\text{ and }B=-1$

$\therefore \dfrac{1}{(1-t)(2-t)}=\dfrac{1}{(1-t)}-\dfrac{1}{(2-t)}$

$\Rightarrow \int{\dfrac{\cos x}{(1-\sin x)(2-\sin x)}}dx=\int{\left\{ \dfrac{1}{1-t}-\dfrac{1}{(2-t)} \right\}}dt=-\log |1-t|+\log |2-t|+C$ Simplifying,

$=\log \left| \dfrac{2-t}{1-t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{2-\sin x}{1-\sin x} \right|+C$

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

Ans:  $\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}$

Let $\dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{Ax+B}{\left( {{x}^{2}}+3 \right)}+\dfrac{Cx+D}{\left( {{x}^{2}}+4 \right)}$

$4{{x}^{2}}+10=(Ax+B)\left( {{x}^{2}}+4 \right)+(Cx+D)\left( {{x}^{2}}+3 \right)$

$4{{x}^{2}}+10=A{{x}^{2}}+4Ax+B{{x}^{2}}+4B+C{{x}^{3}}+3Cx+D{{x}^{2}}+3D$

$4{{x}^{2}}+10=(A+C){{x}^{3}}+(B+D){{x}^{2}}+(4A+3C)x+(4B+3D)$

We obtain the below values by equating the coefficients of ${{\text{x}}^{3}},{{\text{x}}^{2}},\text{x}$and constant term, 

$A+C=0$

$B+D=4$

$4\text{ }A+3\text{ }C=0$

$4\text{ }B+3\text{ }D=10$

On solving these equations, to obtain $\text{A}=0.\text{B}=-2.\text{C}=0,and\,\,\,\text{D}=6$

$\therefore \dfrac{\left( 4{{x}^{2}}+10 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)}$

$\dfrac{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+2 \right)}{\left( {{x}^{2}}+3 \right)\left( {{x}^{2}}+4 \right)}=\left( \dfrac{-2}{\left( {{x}^{2}}+3 \right)}+\dfrac{6}{\left( {{x}^{2}}+4 \right)} \right)$

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

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

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

Simplifying,

$=x+\dfrac{2}{\sqrt{3}}{{\tan }^{-1}}\dfrac{x}{\sqrt{3}}-3{{\tan }^{-1}}\dfrac{x}{2}+C$

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

Ans:  $\dfrac{2x}{\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}+3 \right)}$

Put ${{x}^{2}}-t\to 2xdx-dt$

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

$\text{ Let }\dfrac{1}{(t+1)(t+3)}=\dfrac{A}{(t+1)}+\dfrac{B}{(t+3)}$

$I=A\left( t+3 \right)+B\left( t+1 \right)$

We obtain the below values by equating the coefficients of $\text{t}$and  

 constant, 

                                 $1+B=0$and $3A+B=1$

On solving, we get

$A=\dfrac{1}{2}\text{ and }B=-\dfrac{1}{2}$

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

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

$=\dfrac{1}{2}\log |(t+1)|-\dfrac{1}{2}\log |t+3|+C$

Simplifying,

$=\dfrac{1}{2}\log \left| \dfrac{t+1}{t+3} \right|+C=\dfrac{1}{2}\log \left| \dfrac{{{x}^{2}}+1}{{{x}^{2}}+3} \right|+C$

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

 Ans:  $\dfrac{1}{x\left( {{x}^{4}}-1 \right)}$

Numerator and denominator are multiplied by by ${{\text{x}}^{3}},$we get

$\dfrac{1}{x\left( {{x}^{4}}-1 \right)}=\dfrac{{{x}^{3}}}{{{x}^{4}}\left( {{x}^{4}}-1 \right)}$

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

Consider ${{\text{x}}^{4}}=\text{t}\Rightarrow 4{{\text{x}}^{3}}\text{dx}=\text{dt}$

$\therefore \int{\dfrac{1}{x\left( {{x}^{4}}-1 \right)}}dx=\dfrac{1}{4}\int{\dfrac{dt}{t(t-1)}}$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{(t-1)}$

$1=A(t-1)+Bt\quad \ldots (1)$

We obtain the below values by equating the coefficients of $\text{t}$and constant, 

$A=-1\text{ and }B=1$

$\Rightarrow \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

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

Using the logarithm formula of integration,

$=\dfrac{1}{4}[-\log |t|+\log |t-1|]+C$

Simplifying,

$=\dfrac{1}{4}\log \left| \dfrac{t-1}{t} \right|+C=\dfrac{1}{4}\log \left| \dfrac{{{x}^{4}}-1}{{{x}^{4}}} \right|+C$

21. $\dfrac{1}{{{e}^{x}}-1}$ $\text{[}$ hint: Put ${{e}^{x}}=t$ $\text{]}$

Ans:  $\dfrac{1}{\left( {{e}^{x}}-1 \right)}$

Put ${{\text{e}}^{x}}=\text{t }\Rightarrow {{\text{e}}^{x}}\text{dx}=dt$

$\Rightarrow \int{\dfrac{1}{\left( {{e}^{x}}-1 \right)}}dx=\int{\dfrac{1}{t-1}}\times \dfrac{dt}{t}=\int{\dfrac{1}{t(t-1)}}dt$

$\text{ Let }\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{t-1}$

$1=A\left( t-1 \right)+B\text{ }t$

We obtain the below values by equating the coefficients of t and constant, $A=-1\text{ and }B=1$

$\therefore \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int{\dfrac{1}{t(t-1)}}dt=\log \left| \dfrac{t-1}{t} \right|+C$

Substitute the value of t,

$=\log \left| \dfrac{{{e}^{x}}-1}{{{e}^{x}}} \right|+C$

22. $\int{\dfrac{xdx}{\left( x-1 \right)\left( x-2 \right)}}$ equals

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

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

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

4. $\log \left| \left( x-1 \right)\left( x-2 \right) \right|+C$

 Ans:  Let $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}$

$x=A\left( x-2 \right)+B\left( x-1 \right)$

We obtain the below values by equating the coefficients of $\text{x}$and constant,  $\text{A}=-1\,\,\,and\,\,\text{B}=2$

$\therefore \dfrac{x}{(x-1)(x-2)}=-\dfrac{1}{(x1)}+\dfrac{2}{(x-2)}$

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

Using the logarithm formula of integration,

$=-\log |x-1|+2\log |x-2|+C$

Simplifying,

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

Thus, the right response is B.

23. $\int{\dfrac{dx}{x\left( {{x}^{2}}+1 \right)}}$ equals 

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

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

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

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

Ans: Let $\dfrac{1}{x\left( {{x}^{2}}+1 \right)}-\dfrac{A}{x},\dfrac{Bx+C}{{{x}^{2}}+1}$

$1=A\left( {{x}^{2}}+1 \right)+(Bx+C)x$

We obtain the below values by equating the coefficients of ${{\text{x}}^{2}},\text{x},$     and constant term, 

$A+B=0$, $C=0$$A=1$

On solving these equations, to obtain

$A=1,B=-1,\text{ and }C=0$

$\therefore \dfrac{1}{x\left( {{x}^{2}},1 \right)}=\dfrac{1}{x}+\dfrac{-x}{{{x}^{2}}+1}$

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

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

Thus, the right response is $\text{A}.$

Exercise 7.6

1. $x\sin x$

 Ans:  Let $I=\int{x}\sin xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin \text{x}$and integrating by parts, to obtain  

$I=\int{x}\sin xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }xdx \right\}}dx$

$=x(-\cos x)-\int{1}.(-\cos x)dx$

$=-x\cos x+\sin x+C$

2. $x\sin 3x$

Ans: Let $\text{I}=\int{x}\sin 3xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin 3\text{x}$ and integrating by parts, to obtain  

$I=x\int{\sin }3xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }3xdx \right\}}$

$=x\left( \dfrac{-\cos 3x}{3} \right)-\int{1}\cdot \left( \dfrac{-\cos 3x}{3} \right)dx$

$=\dfrac{-x\cos 3x}{3}+\dfrac{1}{3}\int{\cos }3xdx=\dfrac{-x\cos 3x}{3}+\dfrac{1}{9}\sin 3x+C$

3. ${{x}^{2}}{{e}^{x}}$

Ans:  Let $I=\int{{{x}^{2}}}{{e}^{x}}dx$

Consider  $\text{u}={{\text{x}}^{2}}\,\,and\,\,\text{v}={{\text{e}}^{x}}$

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

$={{x}^{2}}{{e}^{x}}-\int{2}x-{{e}^{x}}dx$

$={{x}^{2}}{{e}^{x}}-2\int{x}\cdot {{e}^{x}}dx$

Again using integration by parts, to obtain

$={{x}^{2}}{{e}^{x}}-2\left[ x\cdot \int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx \right]$

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-\int{{{e}^{x}}}dx \right]$

Simplifying,

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-{{e}^{x}} \right]$

$={{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+2{{e}^{x}}+C$

$={{e}^{x}}\left( {{x}^{2}}-2x+2 \right)+C$

4. $x\log x$

Ans:  Let $I=\int{x}\log xdx$

Consider $\text{u}=\log \text{x}\,\,\,\,and\,\,\,\text{v}=\text{x}$ and integrating by parts, to obtain

$I=\log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log x}{2}\cdot \sqrt{\dfrac{x}{2}}dx=\dfrac{{{x}^{2}}\log x}{2}-\dfrac{{{x}^{2}}}{4}+C$

5. $x\log 2x$

 Ans:  Let $I=\int{x}\log 2xdx$

Consider $u=\log 2x$ and $v=x$and integrating by parts, to obtain

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

$=\log 2x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{2}{2x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log 2x}{2}-\int{\dfrac{x}{2}}dx$

Integrating using the power rule

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

6. ${{x}^{2}}\log x$

 Ans: Let $I=\int{{{x}^{2}}}\log xdx$

Consider $u=\log x$and $v={{x}^{2}}$ and integrating by parts, to obtain

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

$=\log x\left( \dfrac{{{x}^{3}}}{3} \right)-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx$

Integrating using the power rule

$=\dfrac{{{x}^{3}}\log x}{3}-\int{\dfrac{{{x}^{2}}}{3}}dx=\dfrac{{{x}^{3}}\log x}{3}-\dfrac{{{x}^{3}}}{9}+C$

7. $x{{\sin }^{-1}}x$

 Ans: Let $I=\int{x}{{\sin }^{-1}}xdx$

Consider $u={{\sin }^{-1}}x\,\,and\,\,\,v=x$ and integrating by parts, to obtain

$I={{\sin }^{-1}}x\int{x}dx\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{x}dx \right\}}dx$

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

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

Adding and subtracting by 1

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

Simplifying,

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \int{\sqrt{1-{{x}^{2}}}}dx-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right\}$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x-{{\sin }^{-1}}x \right\}+C$

Simplifying,      $=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}x-\dfrac{1}{2}{{\sin }^{-1}}x+C=\dfrac{1}{4}\left( 2{{x}^{2}}-1 \right){{\operatorname{in}}^{-1}}x+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C$

8. $x{{\tan }^{-1}}x$

Ans: Let $I=\int{x}{{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{x}dx \right\}}dx$

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

Adding and subtracting by -1

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

Simplifying,

$=\dfrac{{{x}^{2}}{{\operatorname{lan}}^{-1}}x}{2}-\dfrac{1}{2}\left( x-{{\tan }^{-1}}x \right)+C=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{x}{2}+\dfrac{1}{2}{{\tan }^{-1}}x+C$

9. $x{{\cos }^{-1}}x$

 Ans:Let $I=\int{x}{{\cos }^{-1}}xdx$

 Taking $u={{\cos }^{-1}}x$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\cos }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\cos }^{-1}}x\dfrac{{{x}^{2}}}{2}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx$

Adding and subtracting by -1

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

Simplifying,

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

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

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}{{I}_{1}}-\dfrac{1}{2}{{\cos }^{-1}}x....\left( 1 \right)$

Where ${{I}_{1}}=\int{\sqrt{1-{{x}^{2}}}}dx$

$\Rightarrow {{I}_{1}}=x\int{\sqrt{1-{{x}^{2}}}}-\int{\dfrac{d}{dx}}\sqrt{1-{{x}^{2}}}\int{x}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ \int{\sqrt{1-{{x}^{2}}}}dx+\int{\dfrac{-dx}{\sqrt{1-{{x}^{2}}}}} \right\}$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ {{I}_{1}}+{{\cos }^{-1}}x \right\}\Rightarrow 2{{I}_{1}}=x\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x$

$\therefore {{I}_{1}}=\dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x$

Substituting in (1), we get

$I=\dfrac{x{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\left( \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x \right)-\dfrac{1}{2}{{\cos }^{-1}}x$

Simplifying,

$=\dfrac{\left( 2{{x}^{2}}-1 \right)}{4}{{\cos }^{-1}}x-\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C$

10. ${{\left( {{\sin }^{-1}}x \right)}^{2}}$

Ans: Let $I=\int{{{\left( {{\sin }^{-1}}x \right)}^{2}}}\cdot 1dx$

Consider $\text{u}={{\left( {{\sin }^{-1}}\text{x} \right)}^{2}}$ and $\text{v}=1$and integrating by parts, to obtain

$I=\int{\left( {{\sin }^{-1}}x \right)}\cdot \int{1}dx-\int{\left\{ \dfrac{d}{dx}{{\left( {{\sin }^{-1}}x \right)}^{2}}\cdot \int{1}.dx \right\}}dx$

$={{\left( {{\sin }^{-1}}x \right)}^{2}}x-\int{\dfrac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}\cdot xdx$

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

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

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

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

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+2\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-2x+C$

11. $\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$

Ans:  Let $I=\int{\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

Multiplying and dividing by 2

$I=\dfrac{-1}{2}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\cdot {{\cos }^{-1}}xdx$

Consider  $\text{u}={{\cos }^{-1}}\text{x}$ and $\text{v}=\left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)$and integrating by parts, to obtain

$I=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+\int{2}dx \right]$

Simplifying,

$=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+2x \right]+C$

$=-\left[ \sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+x \right]+C$

12. $x{{\sec }^{2}}x$

 Ans:Let $I=\int{x}{{\sec }^{2}}xdx$

Consider$\text{u}=\text{x}$ and $\text{v}={{\sec }^{2}}\text{x}$ and integrating by parts, to obtain

$I=x\int{{{\sec }^{2}}}xdx-\int{\left\{ \left\{ \dfrac{d}{dx}x \right\}\int{{{\sec }^{2}}}xdx \right\}}dx$

$=x\tan x-\int{1}\cdot \tan xdx$

$=x\tan x+\log |\cos x|+C$

13. ${{\tan }^{-1}}x$

Ans: Let $I=\int{1}\cdot {{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=1$ and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{1}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{1}.dx \right\}}dx={{\tan }^{-1}}xx-\int{\dfrac{1}{1+{{x}^{2}}}}xd$

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

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+C$ 

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

14. $x{{\left( \log x \right)}^{2}}dx$

Ans:  $I=\int{x}{{(\log x)}^{2}}dx$

Consider$u={{(\log x)}^{2}}$ and $v=1$ and integrating by parts, to obtain

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

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \int{2}\log x\cdot \dfrac{1}{x}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

Again using integration by parts, to obtain

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

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \dfrac{{{x}^{2}}}{2}-\log x-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{1}{2}\int{x}dx=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{{{x}^{2}}}{4}+C$

15. $\left( {{x}^{2}}+1 \right)\log x$

 Ans: Let $I=\int{\left( {{x}^{2}}+1 \right)}\log xdx=\int{{{x}^{2}}}\log xdx+\int{\log }xdx$

Let $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}\ldots (1)$

Where,${{I}_{1}}=\int{{{x}^{2}}}\log xdx\,\,\,\,\,\,and\,\,\,{{\text{I}}_{2}}=\int{\log }xdx$

${{I}_{1}}=\int{{{x}^{2}}}\log xdx$ 

Consider$\text{u}=\log \text{x}$ and $v=x^2$ and integrating by parts, to obtain

${{I}_{1}}=\log x-\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{3}}}{3}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{1}{3}\int{{{x}^{2}}}dx$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}\quad \ldots (2)$

${{I}_{2}}=\int{\log }xdx$

Consider $\text{u}=\log \text{x}$ and $\text{v}=1$and integrating by parts, to obtain

${{I}_{2}}=\log x\int{1}.dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{1}.dx \right\}}$

$=\log x\cdot x-\int{\dfrac{1}{x}}xdx$

$=x\log x-x..\left( 3 \right)$

Using equations (2) and (3) in (1), we get

$I=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}+x\log x-x+{{C}_{2}}$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+x\log x-x+\left( {{C}_{1}}+{{C}_{2}} \right)$

$=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C$

16. ${{e}^{x}}\left( \sin x+\cos x \right)$

Ans: Consider$I=\int{{{e}^{x}}}(\sin x+\cos x)dx$

Consider$f(x)=\sin x$

${{f}^{\prime }}(x)=\cos x$

$I=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore I={{e}^{x}}\sin x+C$

17. $\dfrac{x{{e}^{x}}}{{{\left( 1+x \right)}^{2}}}$

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

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

Here, $f(x)=\dfrac{1}{1+x}\quad {{f}^{\prime }}(x)=\dfrac{-1}{{{(1+x)}^{2}}}$

$\Rightarrow \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\dfrac{{{e}^{x}}}{1+x}+C$

18. Integrate the function - ${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

Ans: First simplify –${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

It is known that – 

$1+\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

$1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$

$\therefore {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$={{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \tan \dfrac{x}{2}+1 \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\tan }^{2}}\dfrac{x}{2}+1+2\tan \dfrac{x}{2} \right)$

But, $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right)$

$={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

$\Rightarrow {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

If we say, $f(x)=\tan \dfrac{x}{2}\Rightarrow f'(x)=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$

Thus, we get – $\int{{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)}dx={{e}^{x}}\tan \dfrac{x}{2}+C$

19. Integrate the function - ${{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)$

Ans: Say, $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}$

Suppose, $f(x)=\dfrac{1}{x}\Rightarrow f'(x)=-\dfrac{1}{{{x}^{2}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, we get – $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}=\dfrac{{{e}^{x}}}{x}+C$

20. Integrate the function - $\dfrac{(x-3){{e}^{x}}}{{{(x-1)}^{3}}}$

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

$=\int{{{e}^{x}}\left[ \dfrac{(x-1)}{{{(x-1)}^{3}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx}$

$=\int{{{e}^{x}}\left[ \dfrac{1}{{{(x-1)}^{2}}}-\dfrac{2}{{{(x-1)}^{3}}} \right]dx}$

Suppose, $f(x)=\dfrac{1}{{{(x-1)}^{2}}}\Rightarrow f'(x)=-\dfrac{2}{{{(x-1)}^{3}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\dfrac{{{e}^{x}}}{{{(x-1)}^{2}}}+C}$

21. Integrate the function - ${{e}^{2x}}\sin x$

Ans: Say,  $I=\int{{{e}^{2x}}\sin xdx}$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\sin x\text{   }v={{e}^{2x}}$

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

$=\sin x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( \cos x \right)\dfrac{{{e}^{2x}}}{2} \right]dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\int{\left( {{e}^{2x}}\cos x \right)dx}$

Perform Integration by parts for – $\int{\left( {{e}^{2x}}\cos x \right)dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\cos x \right)\int{{{e}^{2x}}dx} \right]dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( -\sin x \right)\dfrac{{{e}^{2x}}}{2} \right]dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}+\dfrac{1}{2}\int{(\sin x){{e}^{2x}}dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}\left\{ \int{(\sin x){{e}^{2x}}dx} \right\}$

But, $I=\int{{{e}^{2x}}\sin xdx}$

$\Rightarrow I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}I$

$\Rightarrow I+\dfrac{1}{4}I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{2{{e}^{2x}}\sin x}{4}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow 5I={{e}^{2x}}(2\sin x-\cos x)$

Thus, we get – $I=\dfrac{{{e}^{2x}}}{5}(2\sin x-\cos x)+C$.

22. Integrate the function - ${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)$

Ans: Say, $x=\tan \theta \text{ }\Rightarrow \text{dx=se}{{\text{c}}^{2}}\theta d\theta $

$\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)$

But, $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta }$

$\Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)=\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \sin 2\theta  \right)=2\theta $

Therefore, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=\int{2\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

$=2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\theta \text{   }v={{\sec }^{2}}\theta $

$2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }=2\left\{ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left[ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -\int{\left[ \tan \theta  \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -(-\log |\cos \theta |) \right\}+C$

$=2\left\{ \theta \tan \theta +\log |\cos \theta | \right\}+C$

Replace $\theta ={{\tan }^{-1}}x$

$=2\left\{ {{\tan }^{-1}}x\tan ({{\tan }^{-1}}x)+\log |\cos ({{\tan }^{-1}}x)| \right\}+C$

It is known that – ${{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}$

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

$=2\left\{ x{{\tan }^{-1}}x+\log |\dfrac{1}{\sqrt{1+{{x}^{2}}}}| \right\}+C$

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

Here, $\log {{m}^{n}}=n\log m$

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

$=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

Thus, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

23. Choose the correct answer: $\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$ equals

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

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

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

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

Ans: Say, $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

Suppose, $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx$

Rewriting the equation – $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx=\dfrac{1}{3}\int{{{e}^{t}}}dt$

$\Rightarrow I=\dfrac{1}{3}\int{{{e}^{t}}}dt=\dfrac{1}{3}{{e}^{t}}+C$

Replacing $t={{x}^{3}}$

$\Rightarrow I=\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

The correct option is A.

24. Choose the correct answer: $\int{{{e}^{x}}\sec x(1+\tan x)}dx$

1. ${{e}^{x}}\cos x+C$

2. ${{e}^{x}}\sec x+C$

3. ${{e}^{x}}\sin x+C$

4. ${{e}^{x}}\tan x+C$

Ans: Say, $I=\int{{{e}^{x}}\sec x(1+\tan x)}dx$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx$

Suppose, $f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

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

Thus, $I={{e}^{x}}\sec x+C$

The correct option is B.

Exercise- 7.7

1. Integrate the function - $\sqrt{4-{{x}^{2}}}$

Ans: Say, $I=\int{\sqrt{4-{{x}^{2}}}dx}=\int{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

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

$\Rightarrow I=\dfrac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\dfrac{x}{2}+C$

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

2. Integrate the function - $\sqrt{1-4{{x}^{2}}}$

Ans: Say, $I=\int{\sqrt{1-4{{x}^{2}}}dx}=\int{\sqrt{{{1}^{2}}-{{(2x)}^{2}}}dx}$

Let, $2x=t\Rightarrow 2dx=dt$

$x=\dfrac{t}{2}\Rightarrow dx=\dfrac{dt}{2}$

So, we get – $I=\int{\sqrt{{{1}^{2}}-{{\left[ 2(\dfrac{t}{2}) \right]}^{2}}}\dfrac{dt}{2}}=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt}$

$\Rightarrow I=\dfrac{1}{2}\int{\sqrt{{{1}^{2}}-{{\left[ t \right]}^{2}}}dt}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow I=\dfrac{1}{2}\left[ \dfrac{t}{2}\sqrt{1-{{t}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}t \right]+C$

$\Rightarrow I=\left[ \dfrac{t}{4}\sqrt{1-{{t}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}t \right]+C$

Replace – $t=2x$

$\Rightarrow I=\left[ \dfrac{2x}{4}\sqrt{1-{{(2x)}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C$

$\Rightarrow I=\left[ \dfrac{x}{2}\sqrt{1-4{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}2x \right]+C$

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

3. Integrate the function - $\sqrt{{{x}^{2}}+4x+6}$

Ans: First simplify –${{x}^{2}}+4x+6$

${{x}^{2}}+4x+6={{x}^{2}}+4x+4+2$

$=({{x}^{2}}+4x+4)+2={{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x+6}=\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x+6}dx}=\int{\sqrt{{{(x+2)}^{2}}+{{(\sqrt{2})}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

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

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

4. Integrate the function - $\sqrt{{{x}^{2}}+4x+1}$

Ans: First simplify –${{x}^{2}}+4x+1$

${{x}^{2}}+4x+1={{x}^{2}}+4x+4-3$

$=({{x}^{2}}+4x+4)-3={{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x+1}=\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x+1}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(\sqrt{3})}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

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

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

5. Integrate the function - $\sqrt{1-4x-{{x}^{2}}}$

Ans: First simplify –$1-4x-{{x}^{2}}$

$1-4x-{{x}^{2}}=1-4x-{{x}^{2}}-4+4=1+4-({{x}^{2}}+4x+4)$

$=5-({{x}^{2}}+4x+4)={{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}$

$\Rightarrow \sqrt{1-4x-{{x}^{2}}}=\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}$

$\therefore \int{\sqrt{1-4x-{{x}^{2}}}dx}=\int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow \int{\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{{{(\sqrt{5})}^{2}}-{{(x+2)}^{2}}}+\dfrac{{{(\sqrt{5})}^{2}}}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$$=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$

Thus, $\int{\sqrt{1-4x-{{x}^{2}}}dx}=\dfrac{x+2}{2}\sqrt{1-4x-{{x}^{2}}}+\dfrac{5}{2}{{\sin }^{-1}}\dfrac{x+2}{\sqrt{5}}+C$

6. Integrate the function - $\sqrt{{{x}^{2}}+4x+5}$

Ans: First simplify –${{x}^{2}}+4x-5$

${{x}^{2}}+4x-5={{x}^{2}}+4x-5+4-4=({{x}^{2}}+4x+4)-5-4$

$=({{x}^{2}}+4x+4)-9={{(x+2)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+4x-5}=\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}+4x-5}dx}=\int{\sqrt{{{(x+2)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

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

Thus, $\int{\sqrt{{{x}^{2}}+4x-5}dx}=\dfrac{x+2}{2}\sqrt{{{x}^{2}}+4x-5}-\dfrac{9}{2}\log \left| (x+2)+\sqrt{{{x}^{2}}+4x-5} \right|+C$

7. Integrate the function - $\sqrt{1+3x-{{x}^{2}}}$

Ans: First simplify –$1+3x-{{x}^{2}}$

$1+3x-{{x}^{2}}=1-{{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=1+\dfrac{9}{4}-({{x}^{2}}-3x+\dfrac{9}{4})$

$=\dfrac{9+4}{4}-({{x}^{2}}-3x+\dfrac{9}{4})=\left( \dfrac{13}{4} \right)-({{x}^{2}}-3x+\dfrac{9}{4})={{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}$

$\Rightarrow \sqrt{1+3x-{{x}^{2}}}=\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}$

$\therefore \int{\sqrt{1+3x-{{x}^{2}}}dx}=\int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\dfrac{x}{a}+C}$

$\Rightarrow \int{\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x-\dfrac{3}{2} \right)}{2}\sqrt{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}-{{\left( x-\dfrac{3}{2} \right)}^{2}}}+\dfrac{{{\left( \dfrac{\sqrt{13}}{2} \right)}^{2}}}{2}{{\sin }^{-1}}\dfrac{\left( x-\dfrac{3}{2} \right)}{\left( \dfrac{\sqrt{13}}{2} \right)}+C$$=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$

Thus, $\int{\sqrt{1+3x-{{x}^{2}}}dx}=\dfrac{2x-3}{4}\sqrt{1+3x-{{x}^{2}}}+\dfrac{13}{8}{{\sin }^{-1}}\dfrac{2x-3}{\sqrt{13}}+C$

8. Integrate the function - $\sqrt{{{x}^{2}}+3x}$

Ans: First simplify –${{x}^{2}}+3x$

${{x}^{2}}+x={{x}^{2}}+3x+\dfrac{9}{4}-\dfrac{9}{4}=({{x}^{2}}+3x+\dfrac{9}{4})-\dfrac{9}{4}$

$={{\left( x+\dfrac{3}{2} \right)}^{2}}-\left( \dfrac{9}{4} \right)={{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}+3x}=\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}$

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

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}dx}=\dfrac{\left( x+\dfrac{3}{2} \right)}{2}\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}}-\dfrac{{{\left( \dfrac{3}{2} \right)}^{2}}}{2}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}} \right|+C$$=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}+3x}dx}=\dfrac{2x+3}{4}\sqrt{{{x}^{2}}+3x}-\dfrac{9}{8}\log \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x} \right|+C$

9. Integrate the function - $\sqrt{1+\dfrac{{{x}^{2}}}{9}}$

Ans: First simplify –$1+\dfrac{{{x}^{2}}}{9}$

$1+\dfrac{{{x}^{2}}}{9}=\dfrac{1}{9}(9+{{x}^{2}})=\dfrac{1}{9}({{3}^{2}}+{{x}^{2}})$

$\Rightarrow \sqrt{1+\dfrac{{{x}^{2}}}{9}}=\sqrt{\dfrac{1}{9}({{3}^{2}}+{{x}^{2}})}=\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})}$

$\therefore \int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\int{\dfrac{1}{3}\sqrt{({{3}^{2}}+{{x}^{2}})}dx}=\dfrac{1}{3}\int{\sqrt{({{3}^{2}}+{{x}^{2}})}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

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

$\dfrac{x}{6}{\sqrt{x^2+9}}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right| +C$
Thus, $\int{\sqrt{1+\dfrac{{{x}^{2}}}{9}}dx}=\dfrac{x}{6}\sqrt{{{x}^{2}}+9}+\dfrac{3}{2}\log \left| x+\sqrt{{{x}^{2}}+9} \right|+C$

10. Choose the correct answer: $\int{\sqrt{1+{{x}^{2}}}dx}$ is equal to –

1. $\dfrac{x}{2}\sqrt{1+{{x}^{2}}}+\dfrac{1}{2}\log \left| (x+\sqrt{1+{{x}^{2}}}) \right|+C$

2. $\dfrac{2}{3}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

3. $\dfrac{2}{3}x{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}+C$

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

Ans: It is known that – $\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+C}$

Thus,  $\int{\sqrt{{{x}^{2}}+{{1}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}+{{1}^{2}}}+\dfrac{{{1}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}+{{1}^{2}}} \right|+C}$

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

The correct answer is option A.

11. Choose the correct answer: $\int{\sqrt{{{x}^{2}}-8x+7}dx}$ is equal to –

1. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

2. $\dfrac{1}{2}(x+4)\sqrt{{{x}^{2}}-8x+7}+9\log \left| (x+4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

3. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}-3\sqrt{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

4. $\dfrac{1}{2}(x-4)\sqrt{{{x}^{2}}-8x+7}+\dfrac{9}{2}\log \left| (x-4+\sqrt{{{x}^{2}}-8x+7}) \right|+C$

Ans: First simplify –${{x}^{2}}-8x+7$

${{x}^{2}}-8x+7+9-9={{x}^{2}}-8x+16-9=({{x}^{2}}-8x+16)-9$

$={{(x-4)}^{2}}-{{(3)}^{2}}$

$\Rightarrow \sqrt{{{x}^{2}}-8x+7}=\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}$

$\therefore \int{\sqrt{{{x}^{2}}-8x+7}dx}=\int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}$

It is known that – $\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$

$\Rightarrow \int{\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}dx}=\dfrac{(x-4)}{2}\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}}-\dfrac{{{(3)}^{2}}}{2}\log \left| (x-4)+\sqrt{{{(x-4)}^{2}}-{{(3)}^{2}}} \right|+C$$=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

Thus, $\int{\sqrt{{{x}^{2}}-8x+7}dx}=\dfrac{x-4}{2}\sqrt{{{x}^{2}}-8x+7}-\dfrac{9}{2}\log \left| (x-4)+\sqrt{{{x}^{2}}-8x+7} \right|+C$

The correct answer is option D

Exercise- 7.8

1. Evaluate the definite integral as limit of sum – $\int\limits_{a}^{b}{xdx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x\text{ ; }a=a\text{ ; }b=b\text{ ; }h=\dfrac{b-a}{n}$

 $\therefore \int\limits_{a}^{b}{xdx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ a+(a+h)+(a+2h)+...+(a+(n-1)h) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(a+a+a+...+a)}_{n-times}}+(h+2h+...+(n-1)h) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ na+h(1+2+3+...+(n-1)) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ na+h\left( \dfrac{n(n-1)}{2} \right) \right]$

‘n’ is a common factor

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ a+\left( \dfrac{h(n-1)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{h(n-1)}{2} \right) \right]$

Replace the value for ‘h’

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{\left( \dfrac{b-a}{n} \right)(n-1)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)(n-1)}{2n} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\left[ a+\left( \dfrac{(b-a)\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=(b-a)\left[ a+\left( \dfrac{(b-a)\left( 1-0 \right)}{2} \right) \right]$

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

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

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

$=\left( \dfrac{{{b}^{2}}-{{a}^{2}}}{2} \right)$

Thus, $\int\limits_{a}^{b}{xdx}=\dfrac{1}{2}({{b}^{2}}-{{a}^{2}})$

2. Evaluate the definite integral as limit of sum – $\int\limits_{0}^{5}{(x+1)dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x+1\text{ ; }a=0\text{ ; }b=5\text{ ; }h=\dfrac{b-a}{n}=\dfrac{5-0}{n}=\dfrac{5}{n}$

 $\therefore \int\limits_{0}^{5}{(x+1)dx}=(5-0)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(0)+f\left( \dfrac{5}{n} \right)+f\left( \dfrac{10}{n} \right)+...+f\left( \dfrac{(n-1)5}{n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+\left( \dfrac{5}{n}+1 \right)+\left( \dfrac{10}{n}+1 \right)+...+\left( \dfrac{(n-1)5}{n}+1 \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(1+1+1...+1)}_{n-times}}+\left( \dfrac{5}{n} \right)+\left( \dfrac{(2)5}{n} \right)+...+\left( \dfrac{(n-1)5}{n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{5}{n}(1+2+3+...+(n-1) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{5}{n}\left( \dfrac{n(n-1)}{2} \right) \right]$

‘n’ is a common factor

$=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ 1+\left( \dfrac{5(n-1)}{2n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5(n-1)}{2n} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=5\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{5\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=5\left[ 1+\left( \dfrac{5\left( 1-0 \right)}{2} \right) \right]$

$=5\left[ 1+\left( \dfrac{5}{2} \right) \right]$

$=5\left[ \dfrac{7}{2} \right]$

$=\left( \dfrac{35}{2} \right)$

Thus, $\int\limits_{0}^{5}{(x+1)dx}=\dfrac{35}{2}$

3. Evaluate the definite integral as limit of sum – $\int\limits_{2}^{3}{{{x}^{2}}dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)={{x}^{2}}\text{ ; }a=2\text{ ; }b=3\text{ ; }h=\dfrac{3-2}{n}=\dfrac{1}{n}$

 $\therefore \int\limits_{2}^{3}{{{x}^{2}}dx}=(3-2)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(2)+f\left( 2+\dfrac{1}{n} \right)+f\left( 2+\dfrac{2}{n} \right)+...+f\left( 2+\dfrac{(n-1)}{n} \right) \right]$

$=1\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}+{{\left( 2+\dfrac{1}{n} \right)}^{2}}+{{\left( 2+\dfrac{2}{n} \right)}^{2}}+...+{{\left( 2+\dfrac{(n-1)}{n} \right)}^{2}} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}+\left\{ {{2}^{2}}+\dfrac{1}{{{n}^{2}}}+2.2.\dfrac{1}{n} \right\}+\left\{ {{2}^{2}}+\dfrac{{{2}^{2}}}{{{n}^{2}}}+2.2.\dfrac{2}{n} \right\}+...+\left\{ {{2}^{2}}+\dfrac{{{(n-1)}^{2}}}{{{n}^{2}}}+2.2.\dfrac{(n-1)}{n} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{\left\{ {{2}^{2}}+{{2}^{2}}+{{...2}^{2}} \right\}}_{n-times}}+\left\{ \dfrac{1}{{{n}^{2}}}+\dfrac{{{2}^{2}}}{{{n}^{2}}}+...+\dfrac{{{(n-1)}^{2}}}{{{n}^{2}}} \right\}+\left\{ 2.2.\dfrac{1}{n}+2.2.\dfrac{2}{n}+...+2.2.\dfrac{(n-1)}{n} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{2}^{2}}n+\dfrac{1}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+2.2.\dfrac{1}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+\dfrac{4}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

It is also known that the sum of n-squared-terms is – ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{(n-1)}^{2}}=\dfrac{n(n-1)(2n-1)}{6}$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{{{n}^{2}}}\left\{ \dfrac{n(n-1)(2n-1)}{6} \right\}+\dfrac{4}{n}\left\{ \dfrac{n(n-1)}{2} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 4n+\dfrac{1}{n}\left\{ \dfrac{(n-1)(2n-1)}{6} \right\}+4\left\{ \dfrac{(n-1)}{2} \right\} \right]$

Taking $\dfrac{1}{n}$ inside

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\dfrac{1}{{{n}^{2}}}\left\{ \dfrac{(n-1)(2n-1)}{6} \right\}+\dfrac{4}{n}\left\{ \dfrac{(n-1)}{2} \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\left\{ \dfrac{\left( \dfrac{(n-1)}{n} \right)\left( \dfrac{(2n-1)}{n} \right)}{6} \right\}+2\left\{ \left( \dfrac{(n-1)}{n} \right) \right\} \right]$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left[ 4+\left\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 2-\dfrac{1}{n} \right)}{6} \right\}+2\left\{ \left( 1-\dfrac{1}{n} \right) \right\} \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=\left[ 4+\left\{ \dfrac{\left( 1-0 \right)\left( 2-0 \right)}{6} \right\}+2\left\{ 1-0 \right\} \right]$

$=\left[ 4+\left\{ \dfrac{\left( 1 \right)\left( 2 \right)}{6} \right\}+2\left\{ 1 \right\} \right]$

$=\left[ 4+\left\{ \dfrac{1}{3} \right\}+2 \right]$

$=\dfrac{12+1+6}{3}$

$=\dfrac{19}{3}$

Thus, $\int\limits_{2}^{3}{{{x}^{2}}dx}=\dfrac{19}{3}$

4. Evaluate the definite integral as limit of sum – $\int\limits_{1}^{4}{({{x}^{2}}-x)dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Let us say $I={{I}_{1}}-{{I}_{2}}$

With, ${{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}\text{  ;  }{{I}_{2}}=\int\limits_{1}^{4}{xdx}$

Finding ${{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}$

Here, $f(x)={{x}^{2}}\text{ ; }a=1\text{ ; }b=4\text{ ; }h=\dfrac{4-1}{n}=\dfrac{3}{n}$

$\therefore {{I}_{1}}=\int\limits_{1}^{4}{{{x}^{2}}dx}=(4-1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(1)+f\left( 1+\dfrac{3}{n} \right)+f\left( 1+\dfrac{(2)3}{n} \right)+...+f\left( 1+\dfrac{(n-1)3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}+{{\left( 1+\dfrac{3}{n} \right)}^{2}}+{{\left( 1+\dfrac{6}{n} \right)}^{2}}+...+{{\left( 1+\dfrac{3(n-1)}{n} \right)}^{2}} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}+\left\{ {{1}^{2}}+\dfrac{{{3}^{2}}}{{{n}^{2}}}+2.2.\dfrac{3}{n} \right\}+\left\{ {{1}^{2}}+\dfrac{{{2}^{2}}{{.3}^{2}}}{{{n}^{2}}}+2.2.\dfrac{2.3}{n} \right\}+...+\left\{ {{1}^{2}}+\dfrac{{{3}^{2}}{{(n-1)}^{2}}}{{{n}^{2}}}+2.2.\dfrac{(n-1)3}{n} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{\left\{ {{1}^{2}}+{{1}^{2}}+{{...1}^{2}} \right\}}_{n-times}}+\left\{ \dfrac{{{3}^{2}}}{{{n}^{2}}}+\dfrac{{{2}^{2}}{{.3}^{2}}}{{{n}^{2}}}+...+\dfrac{{{(n-1)}^{2}}{{.3}^{2}}}{{{n}^{2}}} \right\}+\left\{ 2.\dfrac{3}{n}+2.\dfrac{2.3}{n}+...+2.\dfrac{(n-1).3}{n} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{1}^{2}}n+\dfrac{{{3}^{2}}}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+2.\dfrac{3}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{9}{{{n}^{2}}}\left\{ {{1}^{2}}+{{2}^{2}}+...+{{(n-1)}^{2}} \right\}+\dfrac{6}{n}\left\{ 1+2+...+(n-1) \right\} \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

It is also known that the sum of n-squared-terms is – ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{(n-1)}^{2}}=\dfrac{n(n-1)(2n-1)}{6}$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{9}{{{n}^{2}}}\left\{ \dfrac{n(n-1)(2n-1)}{6} \right\}+\dfrac{6}{n}\left\{ \dfrac{n(n-1)}{2} \right\} \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}\left\{ \dfrac{(n-1)(2n-1)}{2} \right\}+3(n-1) \right]$

Taking $\dfrac{1}{n}$ inside

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\dfrac{3}{{{n}^{2}}}\left\{ \dfrac{(n-1)(2n-1)}{2} \right\}+\dfrac{3}{n}(n-1) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+3\left\{ \dfrac{\left( \dfrac{(n-1)}{n} \right)\left( \dfrac{(2n-1)}{n} \right)}{2} \right\}+3\left( \dfrac{(n-1)}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+3\left\{ \dfrac{\left( 1-\dfrac{1}{n} \right)\left( 2-\dfrac{1}{n} \right)}{2} \right\}+3\left( 1-\dfrac{1}{n} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=3\left[ 1+3\left\{ \dfrac{\left( 1-0 \right)\left( 2-0 \right)}{2} \right\}+3\left( 1-0 \right) \right]$

$=3\left[ 1+3\left\{ \dfrac{\left( 1 \right)\left( 2 \right)}{2} \right\}+3\left( 1 \right) \right]$

$=3\left[ 1+3+3 \right]$

$=3[7]$

$\therefore {{I}_{1}}=21$

Finding ${{I}_{2}}=\int\limits_{1}^{4}{xdx}$

Here, $f(x)=x\text{ ; }a=1\text{ ; }b=4\text{ ; }h=\dfrac{4-1}{n}=\dfrac{3}{n}$

 $\therefore {{I}_{2}}=\int\limits_{1}^{4}{xdx}=(4-1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(1)+f\left( 1+\dfrac{3}{n} \right)+f\left( 1+\dfrac{2.3}{n} \right)+...+f\left( 1+\dfrac{(n-1).3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 1+\left( 1+\dfrac{3}{n} \right)+\left( 1+\dfrac{2.3}{n} \right)+...+\left( 1+\dfrac{(n-1).3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{(1+1+1+...+1)}_{n-times}}+\left( \dfrac{3}{n}+2.\dfrac{3}{n}+...+(n-1).\dfrac{3}{n} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}(1+2+3+...+(n-1)) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ n+\dfrac{3}{n}\left( \dfrac{n\left( n-1 \right)}{2} \right) \right]$

‘n’ is a common factor

$=3\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}.n\left[ 1+\dfrac{3}{n}\left( \dfrac{(n-1)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\dfrac{3}{n}\left( \dfrac{(n-1)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{3\left( \dfrac{n-1}{n} \right)}{2} \right) \right]$

$=3\underset{n\to \infty }{\mathop{\lim }}\,\left[ 1+\left( \dfrac{3\left( 1-\dfrac{1}{n} \right)}{2} \right) \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

$=3\left[ 1+\left( \dfrac{3\left( 1-0 \right)}{2} \right) \right]$

$=3\left[ 1+\left( \dfrac{3}{2} \right) \right]$

$=3\left( \dfrac{2+3}{2} \right)$

$=3\left( \dfrac{5}{2} \right)$

$=\dfrac{15}{2}$

$\therefore {{I}_{2}}=\dfrac{15}{2}$

Now, $I={{I}_{1}}-{{I}_{2}}$

$\Rightarrow I=21-\dfrac{15}{2}=\dfrac{42-15}{2}=\dfrac{27}{2}$

Thus, $\int\limits_{1}^{4}{({{x}^{2}}-x)dx}=\dfrac{27}{2}$

5. Evaluate the definite integral as limit of sum – $\int\limits_{-1}^{1}{{{e}^{x}}dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f\left( x \right)dx=\left( b-a \right)\underset{n\to \infty }{\mathop{\lim }}\,}\dfrac{1}{n}\left[ f\left( a \right)+f\left( a+h \right)+...+f\left( a+\left( n-1 \right)h \right) \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)={{e}^{x}}\text{ ; }a=-1\text{ ; }b=1\text{ ; }h=\dfrac{1-(-1)}{n}=\dfrac{1+1}{n}=\dfrac{2}{n}$

 $\therefore \int\limits_{-1}^{1}{{{e}^{x}}dx}=(1-(-1))\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(-1)+f\left( -1+\dfrac{2}{n} \right)+f\left( -1+2.\dfrac{2}{n} \right)+...+f\left( -1+(n-1).\dfrac{2}{n} \right) \right]$$=(1+1)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}+{{e}^{\left( -1+\dfrac{2}{n} \right)}}+{{e}^{\left( -1+2.\dfrac{2}{n} \right)}}+...+{{e}^{\left( -1+(n-1).\dfrac{2}{n} \right)}} \right]$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}+{{e}^{-1}}{{e}^{\left( \dfrac{2}{n} \right)}}+{{e}^{-1}}{{e}^{\left( 2.\dfrac{2}{n} \right)}}+...+{{e}^{-1}}{{e}^{\left( (n-1).\dfrac{2}{n} \right)}} \right]$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}\left\{ 1+{{e}^{\left( \dfrac{2}{n} \right)}}+{{e}^{\left( \dfrac{4}{n} \right)}}+...+{{e}^{\left( (n-1).\dfrac{2}{n} \right)}} \right\} \right]$

It is known that the sum of n-terms in Geometric progression with $a=1\text{  ;  }r={{e}^{\dfrac{2}{n}}}$ is – $\dfrac{{{e}^{\dfrac{2n}{n}}}-1}{{{e}^{\dfrac{2}{n}}}-1}=\dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1}$

$=2\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ {{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right\} \right]$

$=2{{e}^{-1}}\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left\{ \dfrac{{{e}^{2}}-1}{{{e}^{\dfrac{2}{n}}}-1} \right\}$

$=2{{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{2}{n}}}-1}{\dfrac{2}{n}} \right]2} \right\}$

It is known that – $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{h}}-1}{h}=1$

$=2{{e}^{-1}}\left\{ \dfrac{{{e}^{2}}-1}{(1)2} \right\}$

$={{e}^{-1}}({{e}^{2}}-1)$

$=({{e}^{1}}-{{e}^{-1}})$

$=e-\dfrac{1}{e}$

Thus,  $\int\limits_{-1}^{1}{{{e}^{x}}dx}=e-\dfrac{1}{e}$

6. Evaluate the definite integral as limit of sum – $\int\limits_{0}^{4}{(x+{{e}^{2x}})dx}$

Ans: The definite integral as a limit of sum is given by –

$\int\limits_{a}^{b}{f(x)dx}=(b-a)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(a)+f(a+h)+...+f(a+(n-1)h \right]$

With, $h=\dfrac{b-a}{n}\to 0\text{ }as\text{ }n\to \infty $

Here, $f(x)=x+{{e}^{2x}}\text{ ; }a=0\text{ ; }b=4\text{ ; }h=\dfrac{4-0}{n}=\dfrac{4}{n}$

$\therefore \int\limits_{0}^{4}{(x+{{e}^{2x}})dx}=(4-0)\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ f(0)+f\left( 0+\dfrac{4}{n} \right)+f\left( 0+2.\dfrac{4}{n} \right)+f\left( 0+3.\dfrac{4}{n} \right)+...+f\left( 0+(n-1)\dfrac{4}{n} \right) \right]$$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( 0+{{e}^{0}} \right)+\left( \dfrac{4}{n}+{{e}^{2.\dfrac{4}{n}}} \right)+\left( 2.\dfrac{4}{n}+{{e}^{2.2.\dfrac{4}{n}}} \right)+...+\left( (n-1)\dfrac{4}{n}+{{e}^{2(n-1)\dfrac{4}{n}}} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( {{e}^{0}}+{{e}^{\dfrac{8}{n}}}+{{e}^{2.\dfrac{8}{n}}}+...+{{e}^{(n-1)\dfrac{8}{n}}} \right)+\left( \dfrac{4}{n}+2.\dfrac{4}{n}+...+(n-1)\dfrac{4}{n} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( 1+{{e}^{\dfrac{8}{n}}}+{{e}^{2.\dfrac{8}{n}}}+...+{{e}^{(n-1)\dfrac{8}{n}}} \right)+\dfrac{4}{n}\left( 1+2+...+(n-1) \right) \right]$

It is known that the sum of n-terms is – $1+2+3+...+(n-1)=\dfrac{n(n-1)}{2}$

Also, the sum of n-terms in Geometric progression with $a=1\text{  ;  }r={{e}^{\dfrac{8}{n}}}$ is – $\dfrac{{{e}^{\dfrac{8n}{n}}}-1}{{{e}^{\dfrac{8}{n}}}-1}=\dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1}$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+\dfrac{4}{n}\left( \dfrac{n(n-1)}{2} \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+2\left( n-1 \right) \right]$

$=4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left( \dfrac{{{e}^{8}}-1}{{{e}^{\dfrac{8}{n}}}-1} \right)+4\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ 2\left( n-1 \right) \right]$

$=4\left\{ \dfrac{{{e}^{8}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{8}{n}}}-1}{\dfrac{8}{n}} \right]8} \right\}+4\underset{n\to \infty }{\mathop{\lim }}\,2\left[ \dfrac{n-1}{n} \right]$

$=4\left\{ \dfrac{{{e}^{8}}-1}{\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{{{e}^{\dfrac{8}{n}}}-1}{\dfrac{8}{n}} \right]8} \right\}+4\underset{n\to \infty }{\mathop{\lim }}\,2\left[ 1-\dfrac{1}{n} \right]$

Here, as  $n\to \infty \Rightarrow \dfrac{1}{n}=0$

And, $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{h}}-1}{h}=1$

$=4\left\{ \dfrac{{{e}^{8}}-1}{(1)8} \right\}+4\left\{ 2\left[ 1-0 \right] \right\}$

$=\left\{ \dfrac{{{e}^{8}}-1}{2} \right\}+4(2)$

$=\dfrac{{{e}^{8}}-1}{2}+8$

$=\dfrac{{e^8 - 1+16}}{2}$

$=\left( \dfrac{{{e}^{8}}+15}{2} \right)$

Thus,  $\int\limits_{0}^{4}{(x+{{e}^{2x}})dx}=\dfrac{{{e}^{8}}+15}{2}$

Exercise-7.9

1. Evaluate the definite integral– $\int\limits_{-1}^{1}{(x+1)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{(x+1)dx}=\dfrac{{{x}^{2}}}{2}+x$

So, $\int\limits_{-1}^{1}{(x+1)dx=}\left[ \dfrac{{{x}^{2}}}{2}+x \right]_{-1}^{1}$

$=\left[ \dfrac{{{1}^{2}}}{2}+1 \right]-\left[ \dfrac{{{(-1)}^{2}}}{2}+(-1) \right]$

$=\left[ \dfrac{1}{2}+1 \right]-\left[ \dfrac{1}{2}-1 \right]$

$=\dfrac{1}{2}+1-\dfrac{1}{2}+1=2$

Thus, $\int\limits_{-1}^{1}{(x+1)dx=}2$

2. Evaluate the definite integral– $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\dfrac{1}{x}dx}=\log \left| x \right|$

So, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\left[ \log \left| x \right| \right]_{2}^{3}$

$=\left[ \log \left| 3 \right| \right]-\left[ \log \left| 2 \right| \right]$

$=\log \dfrac{3}{2}$

Thus, $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\log \dfrac{3}{2}$

3. Evaluate the definite integral– $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=4\left( \dfrac{{{x}^{4}}}{4} \right)-5\left( \dfrac{{{x}^{3}}}{3} \right)+6\left( \dfrac{{{x}^{2}}}{2} \right)+9x$

$={{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x$

So,  $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\left[ {{x}^{4}}-\dfrac{5{{x}^{3}}}{3}+3{{x}^{2}}+9x \right]_{1}^{2}$

$=\left[ {{2}^{4}}-\dfrac{5{{(2)}^{3}}}{3}+3{{(2)}^{2}}+9(2) \right]-\left[ {{1}^{4}}-\dfrac{5{{(1)}^{3}}}{3}+3{{(1)}^{2}}+9(1) \right]$

$=\left[ 16-\dfrac{40}{3}+12+18 \right]-\left[ 1-\dfrac{5}{3}+3+9 \right]$

$=\left[ 46-\dfrac{40}{3} \right]-\left[ 13-\dfrac{5}{3} \right]$

$=46-\dfrac{40}{3}-13+\dfrac{5}{3}$

$=33-\dfrac{35}{3}$

$=\dfrac{99-35}{3}$

$=\dfrac{64}{3}$

Thus, $\int\limits_{1}^{2}{(4{{x}^{3}}-5{{x}^{2}}+6x+9)dx}=\dfrac{64}{3}$

4. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\sin 2xdx}=\dfrac{-\cos 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\left[ \dfrac{-\cos 2x}{2} \right]_{0}^{\dfrac{\pi }{4}}$

$=\left[ \dfrac{-\cos 2\left( \dfrac{\pi }{4} \right)}{2} \right]-\left[ \dfrac{-\cos 0}{2} \right]$

$=\left[ \dfrac{-\cos \left( \dfrac{\pi }{2} \right)}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\left[ \dfrac{0}{2} \right]+\left[ \dfrac{1}{2} \right]$

$=\dfrac{1}{2}$

Thus, $\int\limits_{0}^{\dfrac{\pi }{4}}{\sin 2xdx}=\dfrac{1}{2}$

5. Evaluate the definite integral– $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{\cos 2xdx}=\dfrac{\sin 2x}{2}$

So, $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=\left[ \dfrac{\sin 2x}{2} \right]_{0}^{\dfrac{\pi }{2}}$

$=\left[ \dfrac{\sin 2\left( \dfrac{\pi }{2} \right)}{2} \right]-\left[ \dfrac{\sin 0}{2} \right]$

$=\left[ \dfrac{\sin \pi }{2} \right]+\left[ \dfrac{0}{2} \right]$

$=0+0$

$=0$

Thus,  $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2xdx}=0$

6. Evaluate the definite integral– $\int\limits_{4}^{5}{{{e}^{x}}dx}$

Ans: The second fundamental theorem of integral calculus states that – $\int\limits_{a}^{b}{f(x)dx}=F(b)-F(a)$

 Here, $\int{{{e}^{x}}dx}={{e}^{x}}$

So, $\int\limits_{4}^{5}{{{e}^{x}}dx}=\left[ {{e}^{x}} \right]_{4}^{5}$

$=\left[ {{e}^{5}} \right]-\left[ {{e}^{4}} \right]$

$={{e}^{4}}(e-1)$

Thus, $\int\limits_{4}^{5}{{{e}^{x}}dx}={{e}^{4}}(e-1)$

7. $\int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}$

Ans: We know that,

$\int{\tan xdx}=-\log \left| \cos x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int{\tan x dx} = \left[ -\log \left| \cos x \right| \right]_{0}^{\dfrac{\pi}{4}}$

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$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \cos \dfrac{\pi }{4} \right|+\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \dfrac{1}{\sqrt{2}} \right|+\log \left| 1 \right| \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\dfrac{1}{2}\log 2$

8. $\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}$

Ans: We know that,

$\int{\cos ecxdx}=\log \left| \cos ecx-\cot x \right|+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ecx-\cot x \right| \right]_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \cos ec\dfrac{\pi }{4}-\cot \dfrac{\pi }{4} \right|-\log \left| \cos ec\dfrac{\pi }{6}-\cot \dfrac{\pi }{6} \right| \right]$

$\Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\left[ \log \left| \sqrt{2}-1 \right|-\log \left| 2-\sqrt{3} \right| \right]$

$\therefore \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{4}}{\cos ecxdx}=\log \left( \dfrac{\sqrt{2}-1}{2-\sqrt{3}} \right)$

9. $\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}$

Ans: We know that,

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

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ {{\sin }^{-1}}1-{{\sin }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\left[ \dfrac{\pi }{2}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx}=\dfrac{\pi }{2}$

10. $\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}$

Ans: We know that,

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

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}x \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\left[ \dfrac{\pi }{4}-0 \right]$

$\therefore \int\limits_{0}^{1}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{4}$

11. $\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}$

Ans: We know that,

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

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{x-1}{x+1} \right| \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \left| \dfrac{3-1}{3+1} \right|-\log \left| \dfrac{2-1}{2+1} \right| \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\left[ \log \dfrac{1}{2}-\log \dfrac{1}{3} \right]$

$\therefore \int\limits_{2}^{3}{\dfrac{1}{{{x}^{2}}-1}dx}=\dfrac{1}{2}\log \dfrac{3}{2}$

12. $\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}$

Ans: We know that,

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

$\Rightarrow \int{{{\cos }^{2}}xdx}=\dfrac{x}{2}+\dfrac{\sin 2x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\left[ \dfrac{x}{2}+\dfrac{\sin 2x}{4} \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}-\dfrac{\sin \pi }{2}-0-\dfrac{\sin 0}{2} \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{1}{2}\left[ \dfrac{\pi }{2}+0-0-0 \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$

13. $\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}$

Ans: We know that,

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

$\Rightarrow \int{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)$

Therefore, by second fundamental theorem of calculus

$\int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\left[ \log \left( 1+{{3}^{2}} \right)-\log \left( 1+{{2}^{2}} \right) \right]_{2}^{3}$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\left[ \log 10-\log 5 \right]$

$\Rightarrow \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{2}\log \dfrac{10}{5}$

$\therefore \int\limits_{2}^{3}{\dfrac{xdx}{{{x}^{2}}+1}}=\dfrac{1}{4}\log 2$

14. $\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$

Ans: Solving $\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}$ ,

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

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x+15}{5{{x}^{2}}+1}dx}$

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

$\int{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\int{\dfrac{10x}{5{{x}^{2}}+1}dx}+3\int{\dfrac{1}{5\left( {{x}^{2}}+\dfrac{1}{5} \right)}dx}$

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

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\left\{ \dfrac{1}{5}\log \left( 5+1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\left( \sqrt{5} \right)x \right\}-\left\{ \dfrac{1}{5}\log \left( 1 \right)+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}0 \right\}$

$\therefore \int\limits_{0}^{1}{\dfrac{2x+3}{5{{x}^{2}}+1}dx}=\dfrac{1}{5}\log 6+\dfrac{3}{\sqrt{5}}{{\tan }^{-1}}\sqrt{5}$

15. $\int\limits_{0}^{1}{x{{e}^{{{x}^{2}}}}dx}$

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

Differentiating it we get,

$2xdx=dt$

Therefore, the integral becomes,

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\left[ \dfrac{1}{2}{{e}^{t}} \right]_{0}^{1}$

$\dfrac{1}{2}\int\limits_{0}^{1}{{{e}^{t}}dt}=\dfrac{1}{2}e-\dfrac{1}{2}{{e}^{0}}$

$\dfrac{1}{2}\left( e-1 \right)$

16. $\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}$

Ans: The given integral can be written as

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\int\limits_{1}^{2}{\left\{ 5-\dfrac{20x+15}{{{x}^{2}}+4x+3} \right\}dx}$

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=\left[ 5x \right]_{1}^{2}-\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$                             …(1)

Solving $\int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}$ ,

Let $20x+15=A\dfrac{d}{dx}\left( {{x}^{2}}+4x+3 \right)+B$

Equating the coefficients of $x$ and constant term we get,

$A=10,B=-25$

Let ${{x}^{2}}+4x+3=t$

Differentiating it we get,

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

Therefore, the integral becomes

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}$

$10\int{\dfrac{dt}{t}-25\int{\dfrac{dx}{{{\left( x+2 \right)}^{2}}-{{1}^{2}}}}}=10\log t-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+2-1}{x+2+1} \right) \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\left[ 10\log \left( {{x}^{2}}+4x+3 \right)-25\left[ \dfrac{1}{2}\log \left( \dfrac{x+1}{x+3} \right) \right] \right]_{1}^{2}$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 15-10\log 8-25\left[ \dfrac{1}{2}\log \dfrac{3}{5}-\dfrac{1}{2}\log \dfrac{2}{4} \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=10\log 5+10\log 3-10\log 4-10\log 2-\dfrac{25}{2}\left[ \log 3-\log 5-\log 2+\log 4 \right]$

$\Rightarrow \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log 5-\dfrac{45}{2}\log 4-\dfrac{5}{2}\log 3+\dfrac{5}{2}\log 2$

$\therefore \int\limits_{1}^{2}{\dfrac{20x+15}{{{x}^{2}}+4x+3}dx}=\dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2}$

Substituting it in (1) we get,

$\int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\left[ \dfrac{45}{2}\log \dfrac{5}{4}-\dfrac{5}{2}\log \dfrac{3}{2} \right]$

$\therefore \int\limits_{1}^{2}{\dfrac{5{{x}^{2}}}{{{x}^{2}}+4x+3}dx}=5-\dfrac{5}{2}\left[ 9\log \dfrac{5}{4}-\log \dfrac{3}{2} \right]$

17. $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}$

Ans: We know that,

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan x+\dfrac{{{x}^{4}}}{4}+2x$

Therefore, by second fundamental theorem of calculus

$\int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan x+\dfrac{{{x}^{4}}}{4}+2x \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=\left[ 2\tan \dfrac{\pi }{4}+\dfrac{1}{4}{{\left( \dfrac{\pi }{4} \right)}^{2}}+2\left( \dfrac{\pi }{4} \right)-\left( 2\tan 0+0+0 \right) \right]$

$\Rightarrow \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2\tan \dfrac{\pi }{4}+\dfrac{{{\pi }^{4}}}{{{4}^{5}}}+\dfrac{\pi }{2}$

$\therefore \int{\left( 2{{\sec }^{2}}x+{{x}^{3}}+2 \right)dx}=2+\dfrac{\pi }{2}+\dfrac{{{\pi }^{4}}}{1024}$

18. $\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}$

Ans: We know that,

$\int\limits_{0}^{\pi }{\left( {{\sin }^{2}}\dfrac{x}{2}-{{\cos }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}$

$\Rightarrow -\int\limits_{0}^{\pi }{\left( {{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \right)dx}=-\int\limits_{0}^{\pi }{\cos xdx}$

$\int{\cos xdx}=\sin x+C$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\pi }{\cos xdx}=\sin \pi -sin0$

$\therefore \int\limits_{0}^{\pi }{\cos xdx}=0$

19. $\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}$

Ans: Solving the integral we get,

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x+1}{{{x}^{2}}+4}dx}$

$\int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\int{\dfrac{2x}{{{x}^{2}}+4}dx}+3\int{\dfrac{1}{{{x}^{2}}+4}dx}$

$\therefore \int{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{x}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{x}{2} \right]_{0}^{2}$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=\left[ 3\log \left( {{2}^{2}}+4 \right)+\dfrac{3}{2}{{\tan }^{-1}}\dfrac{2}{2}-3\log \left( {{0}^{2}}+4 \right)-\dfrac{3}{2}{{\tan }^{-1}}\dfrac{0}{2} \right]$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}{{\tan }^{-1}}1-3\log 4-\dfrac{3}{2}{{\tan }^{-1}}0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 8+\dfrac{3}{2}\left( \dfrac{\pi }{4} \right)-3\log 4-0$

$\Rightarrow \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log \dfrac{8}{4}+\dfrac{3\pi }{8}$

$\therefore \int\limits_{0}^{2}{\dfrac{6x+3}{{{x}^{2}}+4}dx}=3\log 2+\dfrac{3\pi }{8}$

20. $\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}$

Ans: Solving the integral we get,

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x\int{{{e}^{x}}dx}-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{{{e}^{x}}dx} \right\}dx}+\left\{ \dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi }{4}} \right\}$

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-\int{{{e}^{x}}dx}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=x{{e}^{x}}-{{e}^{x}}-\dfrac{4}{\pi }\cos \dfrac{x}{4}$

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1{{e}^{1}}-{{e}^{1}}-\dfrac{4}{\pi }\cos \dfrac{\pi }{4} \right)-\left( 0{{e}^{0}}-{{e}^{0}}-\dfrac{4}{\pi }\cos 0 \right)$

$\therefore \int\limits_{0}^{1}{\left( x{{e}^{x}}+\sin \dfrac{\pi x}{4} \right)dx}=\left( 1+\dfrac{4}{\pi }-\dfrac{2\sqrt{2}}{\pi } \right)$

21. $\int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}$

1. $\dfrac{\pi }{3}$

2. $\dfrac{2\pi }{3}$

3. $\dfrac{\pi }{6}$

4. $\dfrac{\pi }{12}$

Ans: Solving the integral we get,

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

Therefore, by second fundamental theorem of calculus