CBSE Class 12 Maths Chapter 7 Integrals – NCERT Solutions Exercise 7.1 [2025-26]

Exercise 7.1

1. Find an anti-derivative (or integral) of the following functions by the method of inspection.

sin $2x$

Ans:  We use the method of inspection as follows:

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

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

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

2. Find an anti-derivative (or integral) of the following functions by the method of inspection. 

cos $3x$

Ans: We use the method of inspection as follows:

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

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

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

3. Find an anti-derivative (or integral) of the following functions by the method of inspection. 

${{e}^{2x}}$

Ans: We use the method of inspection as follows:

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

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

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

4. Find an anti-derivative (or integral) of the following functions by the method of inspection.

${{\left( ax+b \right)}^{2}}$

Ans: We use the method of inspection as follows:

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

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

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

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

5. Find an anti-derivative (or integral) of the following functions by the method of inspection. 

sin $2x-4{{e}^{3x}}$

Ans: We use the method of inspection as follows:

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

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

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( \frac{{{e}^{3x}}}{3} \right)+x+C $ 

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

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

Ans: 

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

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

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

$=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+D $

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( \frac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C $ 

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

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

Ans:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We obtain, on dividing:

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

$=\int{{{x}^{2}}dx}+\int{1.dx} $ 

$=\frac{{{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}^{\frac{3}{2}}} \right)dx} $ 

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

$=\frac{2}{3}{{x}^{\frac{3}{2}}}-\frac{2}{5}{{x}^{\frac{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( {{x}^{\frac{5}{2}}}+2{{x}^{\frac{3}{2}}}+3{{x}^{\frac{1}{2}}} \right)} $  $=3\int{{{x}^{\frac{5}{2}}}dx+2\int{{{x}^{\frac{3}{2}}}dx+3\int{{{x}^{\frac{1}{2}}}}dx}} $  $=\frac{6}{7}{{x}^{\frac{7}{2}}}+\frac{4}{5}{{x}^{\frac{5}{2}}}+2{{x}^{\frac{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}} $ 

$=\frac{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}^{\frac{1}{2}}}}dx}} $ 

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

 $=\frac{2}{3}{{x}^{3}}+3\cos x+\frac{10}{3}{{x}^{\frac{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. $\int{\frac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

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

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

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

$=\int{{{\tan }^{2}}xdx} $ 

$=\int{{{\sec }^{2}}xdx}-\int{1dx} $ 

$=\tan x-x+C $

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

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

$=\int{\left( \frac{2}{{{\cos }^{2}}x}-\frac{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 $\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)$ equals

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

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

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

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

Ans:

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

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

Thus, the correct answer is C.

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

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

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

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

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

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

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

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

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

$Also, $ 

$f\left( 2 \right)=0 $ 

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

$\Rightarrow 16+\frac{1}{8}+C=0 $ 

$\Rightarrow C=\frac{-129}{8} $ 

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

Thus, the correct answer is A.

Conclusion

In conclusion, Exercise 7.1 Class 12 Maths is a crucial stepping stone in mastering the concept of integrals, a fundamental aspect of calculus. By focusing on the basics of indefinite integrals and their standard formulas, Ex 7.1 Class 12 helps students build a strong foundation. Completing these problems not only enhances problem-solving skills but also prepares students for more advanced integration techniques in subsequent exercises and chapters.

Practice and ensure a thorough understanding, as the concepts learned here are integral to various applications in mathematics and beyond. Keep practicing, and students will find themselves well-prepared for more complex calculus challenges ahead!

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