NCERT Solutions for CBSE Class 12 Maths 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

$\int{\tan x dx} = \left[ -\log \left| \cos x \right| \right]_{0}^{\dfrac{\pi}{4}}$

(image will be uploaded soon)

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \cos \dfrac{\pi }{4} \right|+\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\left[ -\log \left| \dfrac{1}{\sqrt{2}} \right|+\log \left| 1 \right| \right]$

$\therefore \int\limits_{0}^{\dfrac{\pi }{4}}{\tan xdx}=\dfrac{1}{2}\log 2$

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

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

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

$\therefore \int\limits_{1}^{\sqrt{3}}{\dfrac{1}{1+{{x}^{2}}}dx}=\dfrac{\pi }{12}$

Thus, the correct option is (D)

22. $\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}$

1. $\dfrac{\pi }{6}$

2. $\dfrac{\pi }{12}$

3. $\dfrac{\pi }{24}$

4. $\dfrac{\pi }{4}$

Ans: Solving the integral we get,

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

Let $3x=t$ ,

Differentiating it we get,

$3dx=dt$

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

Therefore, by second fundamental theorem of calculus

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}{{\tan }^{-1}}\left( \dfrac{3}{2}\cdot \dfrac{2}{3} \right)-\dfrac{1}{6}{{\tan }^{-1}}0$

$\int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{1}{6}\cdot \dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{2}{3}}{\dfrac{1}{4+9{{x}^{2}}}dx}=\dfrac{\pi }{24}$

Thus, the correct answer is (C).

Exercise 7.10

Solve the following integrals.

1. $\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}$

Ans: Let ${{x}^{2}}+1$

Differentiating $2xdx=dt$ ,

Therefore, the integral becomes

$\int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\int\limits_{1}^{2}{\dfrac{dt}{t}}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log \left| t \right| \right]_{1}^{2}$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\left[ \log 2-\log 1 \right]$

$\Rightarrow \int\limits_{0}^{1}{\dfrac{x}{{{x}^{2}}+1}dx}=\dfrac{1}{2}\log 2$

2. $\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }$

Ans: The integral can be written as:

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{4}}\phi \cos \phi d\phi }$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$

Let $\sin \phi =t$

Differentiating it we get,

$\cos \phi d\phi =dt$

Therefore, the integral becomes

$\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}{{\left( 1-{{t}^{2}} \right)}^{2}}dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\sqrt{t}\left( 1+{{t}^{4}}-2{{t}^{2}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\int\limits_{0}^{1}{\left( {{t}^{\dfrac{1}{2}}}+{{t}^{\dfrac{9}{2}}}-2{{t}^{\dfrac{5}{2}}} \right)dt}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\left[ \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+\dfrac{{{t}^{\dfrac{11}{2}}}}{\dfrac{11}{2}}-\dfrac{2{{t}^{\dfrac{7}{2}}}}{\dfrac{7}{2}} \right]_{0}^{1}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{2}{3}+\dfrac{2}{11}-\dfrac{4}{7}$

$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{154+42-132}{231}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{\sin \phi }{{\cos }^{5}}\phi d\phi }=\dfrac{64}{231}$

3. $\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}$

Ans: Let $x=\tan \theta $

Differentiating it we get,

$dx={{\sec }^{2}}\theta d\theta $

Therefore, the integral becomes

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\left( \tan \theta  \right)}^{2}}} \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\sin }^{-1}}\left( \sin 2\theta  \right){{\sec }^{2}}\theta d\theta }$

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\int\limits_{0}^{\dfrac{\pi }{4}}{\theta {{\sec }^{2}}\theta d\theta }$

Let $u=\theta $

And $v={{\sec }^{2}}\theta $

Using integration by parts we get,

$\int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left\{ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right\}d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\int{\tan \theta d\theta } \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \theta \tan \theta -\log \left| \cos \theta  \right| \right]_{0}^{\dfrac{\pi }{4}}$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}\tan \dfrac{\pi }{4}-\log \left| \cos \dfrac{\pi }{4} \right|-\log \left| \cos 0 \right| \right]$

$\Rightarrow \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=2\left[ \dfrac{\pi }{4}-\log \left| \dfrac{1}{\sqrt{2}} \right|-\log 1 \right]$

$\therefore \int\limits_{0}^{1}{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)dx}=\dfrac{\pi }{2}-\log 2$

4. $\int\limits_{0}^{2}{x\sqrt{x+2}dx}$ 

(Put $x+2={{t}^{2}}$)

Ans: