NCERT Solutions for Class 12 Maths Chapter 7 Integrals

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

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

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

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

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

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

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

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

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

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

• Miscellaneous Exercise: This exercise covers various miscellaneous integrals and teaches students how to solve problems using different methods. This exercise contains a collection of problems that require students to apply the concepts they have learned in the previous exercises.

Access NCERT Solutions for Class 12 Mathematics  Chapter 7- Integrals

Exercise 7.1

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

Ans:  We use the method of inspection as follows:

${\displaystyle \frac{d}{dx}}(\cos 2x)=-2\sin 2x\Rightarrow -{\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dx}}(\cos 2x)$

$\therefore \sin 2x={\displaystyle \frac{d}{dx}}(-{\displaystyle \frac{1}{2}}\cos 2x)$

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

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:

${\displaystyle \frac{d}{dx}}(\sin 3x)=3\cos 3x\Rightarrow {\displaystyle \frac{1}{3}}{\displaystyle \frac{d}{dx}}(\sin 3x)$

 $\therefore \cos 3x={\displaystyle \frac{d}{dx}}({\displaystyle \frac{1}{3}}\sin 3x)$

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

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

Ans: We use the method of inspection as follows:

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

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

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

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

Ans: We use the method of inspection as follows:

  ${\displaystyle \frac{d}{dx}}{(ax+b)}^3=3a{(ax+b)}^2$ 

$\Rightarrow {(ax+b)}^2={\displaystyle \frac{1}{3a}}{\displaystyle \frac{d}{dx}}{(ax+b)}^3$ 

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

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

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

Ans: We use the method of inspection as follows:

${\displaystyle \frac{d}{dx}}(-{\displaystyle \frac{1}{2}}\cos 2x-{\displaystyle \frac{4}{3}}{e}^{3x})=(\sin 2x-4e^{3x})$

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

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

Ans: 

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

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

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

$={\displaystyle \frac{4}{3}}{e}^{3x}+x+C$ 

7. $\int x^2(1-{\displaystyle \frac{1}{x^2}})dx$

Ans: $\int x^2(1-{\displaystyle \frac{1}{x^2}})dx$

$=\int (x^2-1)dx$

$={\displaystyle \frac{x^3}{3}}-x+C$

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

Ans:$\int (a{x}^2+bx+c)dx$ 

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

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

9. $\int (2x^2+e^x)dx$

Ans:  $\int (2x^2+e^x)dx$

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

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

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

10. $\int {(\sqrt{x}-{\displaystyle \frac{1}{\sqrt{x}}})}^{2}dx$

Ans:$\int {(\sqrt{x}-{\displaystyle \frac{1}{\sqrt{x}}})}^{2}dx$

$=\int (x+{\displaystyle \frac{1}{x}}-2)dx$

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

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

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

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

$$\int (x+5-4x^{-2})dx$$

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

$$={\displaystyle \frac{x^2}{2}}+5x+{\displaystyle \frac{4}{x}}+C$$

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

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

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

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

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

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

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

We obtain, on dividing:

$=\int (x^2+1)dx$

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

$={\displaystyle \frac{x^3}{3}}+x+C$

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

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

 $=\int (\sqrt{x}-x^{{\displaystyle \frac{3}{2}}})dx$

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

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

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

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

$=3\int (2x^{{\displaystyle \frac{5}{2}}}+2x^{{\displaystyle \frac{3}{2}}}+3x^{{\displaystyle \frac{1}{2}}})$

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

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

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

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

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

$$={\displaystyle \frac{2x^2}{2}}-3(\sin x)+e^x+C$$

$$=x^2-3\sin x+e^x+C$$

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

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

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

= $${\displaystyle \frac{2x^3}{3}}-3(-\cos x)+5\left({\displaystyle \frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}}\right)+C$$ 

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

18. $\int \sec x(\sec x+\tan x)dx$

Ans: $\int \sec x(\sec x+\tan x)dx$

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

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

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

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

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

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

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

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

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

$=\tan x-x+C$

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

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

$=\int ({\displaystyle \frac{2}{{\cos }^{2}x}}-{\displaystyle \frac{3\sin x}{{\cos }^{2}x}})dx$

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

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

21. The anti – derivative of $(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}})$ equals

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

Ans: 

 $(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}})$

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

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

 Thus, the correct answer is C.

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

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

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

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

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

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

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

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

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

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

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

Also, 

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

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

$\Rightarrow 16+{\displaystyle \frac{1}{8}}+C=0$ 

$\Rightarrow C={\displaystyle \frac{-129}{8}}$

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

 Thus, the correct answer is A.

Exercise 7.2

1. ${\displaystyle \frac{2x}{1+x^2}}$

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

$\therefore 2xdx=dt$

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

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

$=\log |1+x^2|+C$

$=\log (1+x^2)+C$

2. ${\displaystyle \frac{{(\log x)}^2}{x}}$

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

$\therefore {\displaystyle \frac{1}{x}}dx=dt$

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

$={\displaystyle \frac{t^3}{3}}+C$

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

3. ${\displaystyle \frac{1}{x+x\log x}}$

Ans: ${\displaystyle \frac{1}{x+x\log x}}={\displaystyle \frac{1}{x(1+\log x)}}$

Substitute $1+\log x=t$

$\therefore {\displaystyle \frac{1}{x}}dx=dt$

$\Rightarrow \int {\displaystyle \frac{1}{x(1+\log x)}}dx=\int {\displaystyle \frac{1}{t}}dt$

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

$=\log |1+\log x|+C$          

4. $Sinx.\sin (\cos x)$

Ans: $\mathrm{Sin}x.\sin (\cos x)$

 $Put,\cos x=t$ 

 $\therefore -\sin xdx=dt$ 

 $\Rightarrow \int \sin x.\sin (\cos x)dx=-\int \sin tdt$

 $=-[-\cos t]+C$ 

 $=\cos t+C$ 

 $=\cos (\cos x)+C$

5. $\sin (ax+b)\cos (ax+b)$

Ans: $\sin (ax+b)\cos (ax+b)={\displaystyle \frac{2\sin (ax+b)\cos (ax+b)}{2}}={\displaystyle \frac{\sin 2(ax+b)}{2}}$ 

Substitute $2(ax+b)=t$

$\therefore 2adx=dt$

$\Rightarrow \int {\displaystyle \frac{\sin 2(ax+b)}{2}}dx={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{\sin t}{2a}}dt$

$={\displaystyle \frac{1}{4a}}[-\cos t]+C$

$={\displaystyle \frac{-1}{4a}}\cos 2(ax+b)+C$

6. $\sqrt{ax+b}$

Ans: Substitute $ax+b=t$

$\Rightarrow adx=dt$

$\therefore dx={\displaystyle \frac{1}{a}}dt$

 $\Rightarrow \int {(ax+b)}^{{\displaystyle \frac{1}{2}}}dx={\displaystyle \frac{1}{a}}\int t^{{\displaystyle \frac{1}{2}}}dt$ 

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

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

Ans: Substitute $x+2=t$

$\therefore dx=dt$

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

$=\int (t^{{\displaystyle \frac{3}{2}}}-2t^{{\displaystyle \frac{1}{2}}})dt$

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

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

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

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

$\therefore 4xdx=dt$

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

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

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

$={\displaystyle \frac{1}{6}}{(1+2x^2)}^{{\displaystyle \frac{3}{2}}}+C$

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

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

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

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

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

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

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

10. ${\displaystyle \frac{1}{x-\sqrt{x}}}$

Ans: ${\displaystyle \frac{1}{x-\sqrt{x}}}={\displaystyle \frac{1}{\sqrt{x}(\sqrt{x}-1)}}$

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

$\Rightarrow \int {\displaystyle \frac{1}{\sqrt{x}(\sqrt{x}-1)}}dx=\int {\displaystyle \frac{2}{t}}dt$ 

$=2\log \left|t\right|+C$ 

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

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

Ans: Substitute $x+4=t$

$\therefore dx=dt$

$\int {\displaystyle \frac{x}{\sqrt{x+4}}}dx=\int {\displaystyle \frac{(t-4)}{\sqrt{t}}}dt=\int (\sqrt{t}-{\displaystyle \frac{4}{\sqrt{t}}})dt$

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

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

 $={\displaystyle \frac{2}{3}}{t}^{{\displaystyle \frac{1}{2}}}(t-12)+C$

$={\displaystyle \frac{2}{3}}\sqrt{x+4}(x-8)+C$

12. ${(x^3-1)}^{{\displaystyle \frac{1}{3}}}{x}^5$

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

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

$\Rightarrow \int {(x^3-1)}^{{\displaystyle \frac{1}{3}}}{x}^{5}dx=\int {(x^3-1)}^{{\displaystyle \frac{1}{3}}}{x}^3.x^{2}dx$

$=\int t^{{\displaystyle \frac{1}{3}}}(t+1){\displaystyle \frac{dt}{3}}={\displaystyle \frac{1}{3}}\int (t^{{\displaystyle \frac{4}{3}}}+t^{{\displaystyle \frac{1}{3}}})dt$

$={\displaystyle \frac{1}{3}}[{\displaystyle \frac{3}{7}}{t}^{{\displaystyle \frac{7}{3}}}+{\displaystyle \frac{3}{4}}{t}^{{\displaystyle \frac{4}{3}}}]+C$

$={\displaystyle \frac{1}{7}}{(x^3-1)}^{{\displaystyle \frac{7}{3}}}+{\displaystyle \frac{1}{4}}{(x^3-1)}^{{\displaystyle \frac{4}{3}}}+C$

13. ${\displaystyle \frac{x^2}{{(2+3x^3)}^3}}$

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

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

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

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

$={\displaystyle \frac{-1}{18{(2+3x^3)}^2}}+C$ 

14. ${\displaystyle \frac{1}{x{(\log x)}^m}},x>0$

Ans: Substitute $\log x=t$

${\displaystyle \frac{1}{x}}dx=dt$

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

$={\displaystyle \frac{{(\log x)}^{1-m}}{(1-m)}}+C$

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

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

$\therefore -8xdx=dt$

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

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

$={\displaystyle \frac{-1}{8}}\log |9-4x^2|+C$

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

Ans: Substitute $2x+3=t$

$\therefore 2dx=dt$

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

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

$={\displaystyle \frac{1}{2}}{e}^{(2x+3)}+C$

17. ${\displaystyle \frac{x}{e^{x^2}}}$

Ans: Substitute $x^2=t$

$\therefore 2xdx=dt$

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

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

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

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

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

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

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

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

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

 $=e^t+C$ 

$=e^{{\tan }^{-1}x}+C$ 

19. Integrate $${\displaystyle \frac{e^{2x}-1}{e^{2x}+1}}$$

Ans:

Dividing the given function’s numerator and denominator by  $e^x$ , we obtain,

$${\displaystyle \frac{{\displaystyle \frac{(e^{2x}-1)}{e^x}}}{{\displaystyle \frac{(e^{2x}+1)}{e^x}}}}={\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}}$$

Let $$e^x+e^{-x}=t$$

$$(e^x-e^{-x})dx=dt$$

$$\Rightarrow \int {\displaystyle \frac{e^{2x}-1}{e^{2x}+1}}dx=\int {\displaystyle \frac{e^x-e^{-x}}{e^x+e^{-x}}}dx$$

$$=\int {\displaystyle \frac{dt}{t}}$$

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

$$=\log |e^x-e^{-x}|+C$$

where C is an arbitrary constant.

20. Solve the following: ${\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}$.

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

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

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

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

Integration of given expression is

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

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

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

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

$\therefore \int {\displaystyle \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}}={\displaystyle \frac{1}{2}}\log |e^{2x}+e^{-2x}|+C$