CUET Mathematics - Calculus - Integrals | Free Questions - Page 2

Q1:

2026: 29 May Shift 1

Calculus

>

Integrals

Medium

Core

$\displaystyle\int_1^3 \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}dx$ is equal to:

-1

2

1

0

Correct Answer

Option 3

Correct Answer

Explanation

Q2:

2026: 29 May Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{-2}^{2}\cos^4x\sin^5x\,dx$	(I) $\pi$
(B) $\displaystyle\int_{-1}^{1}\lvert x \rvert dx$	(II) $\dfrac{\pi}{4}$
(C) $\displaystyle\int_0^{\pi/2}\cos^2x\,dx$	(III) 0
(D) $\displaystyle\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+\tan^5x+1)dx$	(IV) 1

Choose the correct answer from the options given below:

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

(A) - (I), (B) - (IV), (C) - (II), (D) - (III)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

Correct Answer

Option 1

Correct Answer

Explanation

Q3:

2026: 29 May Shift 1

Calculus

>

Integrals

Easy

Applied

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{-1}^{1}(2x^5+3x^3+4x+1)dx=$	(I) 2
(B) $\displaystyle\int_{-2}^{2}(4x^3+2)dx=$	(II) 12
(C) $\displaystyle\int_{-1}^{1}x^3dx=$	(III) 8
(D) $\displaystyle\int_{-3}^{3}(3x^5+2)dx=$	(IV) 0

Choose the correct answer from the options given below:

(A) - (III), (B) - (I), (C) - (II), (D) - (IV)

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

(A) - (I), (B) - (III), (C) - (IV), (D) - (II)

Correct Answer

Option 4

Correct Answer

Explanation

Q4:

2026: 26 May Shift 2

Calculus

>

Integrals

Easy

Common

The value of $\displaystyle\int_0^1 xe^{x^2}\,dx$ is equal to :

$e+1$

$\dfrac{e+1}{2}$

$e-1$

$\dfrac{e-1}{2}$

Correct Answer

Option 4

Correct Answer

Explanation

Q5:

2026: 26 May Shift 2

Calculus

>

Integrals

Easy

Common

$\displaystyle\int x\sqrt{x^2-1}\,dx$ is equal to : ( where C is an arbitrary constant )

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

$\dfrac{2}{3}(x^2-1)^{3/2}+C$

$\dfrac{1}{3}(x^2-1)^{3/2}+C$

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

Correct Answer

Option 3

Correct Answer

Explanation

Q6:

2026: 26 May Shift 2

Calculus

>

Integrals

Hard

Core

Value of $\displaystyle\int_{-1}^{1} \dfrac{x^5+|x|+1}{x^2+2|x|+1}\,dx$ is equal to :

0

$\log_e 2$

$\log_e 4$

$\dfrac{1}{2}\log_e 2$

Correct Answer

Option 3

Correct Answer

Explanation

Q7:

2026: 26 May Shift 2

Calculus

>

Integrals

Easy

Core

$\displaystyle\int \dfrac{1-\sin x}{\cos^2 x}\,dx$ is equal to:

(where $c$ is an arbitrary constant)

$\cot x - \cosec x + c$

$\cosec x - \cot x + c$

$\sec x - \tan x + c$

$\tan x - \sec x + c$

Correct Answer

Option 4

Correct Answer

Explanation

Q8:

2026: 26 May Shift 2

Calculus

>

Integrals

Medium

Core

$\displaystyle\int \dfrac{1}{e^x-1}\,dx$ is equal to:

(where C is an arbitrary constant)

$-(e^x-1)^{-2}(e^x)+c$

$\log_e\left|\dfrac{e^x}{e^x-1}\right|+c$

$\log_e\left|\dfrac{e^x-1}{e^x}\right|+c$

$e^x+c$

Correct Answer

Option 3

Correct Answer

Explanation

Q9:

2026: 26 May Shift 2

Calculus

>

Integrals

Medium

Applied

$\displaystyle\int \dfrac{x}{x-\sqrt{x^2-1}}\,dx$ is equal to :

$x^3 + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^4}{3} + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^3}{3} - \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

$\dfrac{x^3}{3} + \dfrac{(x^2-1)^{3/2}}{3} + C$ ,where $C$ is an arbitrary constant.

Correct Answer

Option 4

Correct Answer

Explanation

Q10:

2026: 26 May Shift 2

Calculus

>

Integrals

Medium

Applied

$\displaystyle\int \sqrt[3]{x}(x^3+4x^2+7)\,dx$ is equal to-

$3\left[13x^{\frac{3}{13}}+\dfrac{5}{2}x^{\frac{3}{10}}+\dfrac{4}{7}x^{\frac{3}{4}}\right]+c$ ,where $C$ is an arbitrary constant.

$3\left[\dfrac{1}{13}x^{\frac{13}{3}}+\dfrac{2}{5}x^{\frac{10}{3}}+\dfrac{7}{4}x^{\frac{4}{3}}\right]+c$ ,where $C$ is an arbitrary constant.

$\dfrac{1}{3}\left[\dfrac{1}{13}x^{13/3}+\dfrac{5}{2}x^{10/3}+\dfrac{7}{4}x^{4/3}\right]+c$ ,where $C$ is an arbitrary constant.

$3\left[13x^{\frac{3}{13}}+\dfrac{2}{5}x^{\frac{3}{10}}+\dfrac{4}{7}x^{\frac{3}{4}}\right]+c$ ,where $C$ is an arbitrary constant.

Correct Answer

Option 2

Correct Answer

Explanation

Q11:

2026: 25 May Shift 1

Calculus

>

Integrals

Easy

Common

If $\dfrac{d}{dx}[f(x)] = 5x^4 - \dfrac{2}{x^3}$ such that $f\left(-\dfrac{1}{2}\right) = 0$ , then $f(x)$ is equal to

$x^5 + \dfrac{1}{x^2} - \dfrac{127}{32}$

$x^4 - \dfrac{1}{x^3} + \dfrac{127}{32}$

$x^5 + \dfrac{1}{x^2} + \dfrac{127}{32}$

$x^4 + \dfrac{1}{x^3} - \dfrac{127}{32}$

Correct Answer

Option 1

Correct Answer

Explanation

Q12:

2026: 25 May Shift 1

Calculus

>

Integrals

Medium

Common

$\displaystyle\int e^{x\log_e 7} e^x\, dx$ is equal to :

[where C is an arbitrary constant]

$\dfrac{7^x}{1+\log_e 7} + C$

$\dfrac{7^x e^x}{1+\log_e 7} + C$

$7^x e^x + C$

$\dfrac{e^x}{1+\log 7} + C$

Correct Answer

Option 2

Correct Answer

Explanation

Q13:

2026: 25 May Shift 1

Calculus

>

Integrals

Hard

Common

$\displaystyle\int \dfrac{1}{(x-1)^{3/4}(x+2)^{5/4}}\, dx$ is equal to :

[where C is an arbitrary constant]

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

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

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

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

Correct Answer

Option 2

Correct Answer

Explanation

Q14:

2026: 25 May Shift 1

Calculus

>

Integrals

Medium

Core

$\displaystyle\int_{-2}^{2} \dfrac{x^2}{1+5^x}\, dx$ is equal to:

8

3/8

8/3

2/3

Correct Answer

Option 3

Correct Answer

Explanation

Q15:

2026: 25 May Shift 1

Calculus

>

Integrals

Medium

Core

If $\displaystyle\int \dfrac{\sin^4 x}{\cos^8 x}\, dx = a\tan^7 x + b\tan^5 x + C$ , where 'C' is an arbitrary constant then

$5a = -7b$

$5b = 7a$

$5b = -7a$

$7b = 5a$

Correct Answer

Option 2

Correct Answer

Explanation

Q16:

2026: 25 May Shift 1

Calculus

>

Integrals

Easy

Core

$\displaystyle\int_{-\pi/3}^{\pi/3} 2x^5\cos^4 x\, dx$ equals

0

-1

1

2

Correct Answer

Option 1

Correct Answer

Explanation

Q17:

2026: 25 May Shift 1

Calculus

>

Integrals

Medium

Applied

Integral $I = \displaystyle\int \dfrac{e^{3\log x} - 3x^4}{e^{2\log x} + x^2}\, dx$ (where $x>0$ ) is equal to

$\dfrac{1}{2}x - \dfrac{3}{2}x^2 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{4}x - \dfrac{1}{2}x^2 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{2}x^2 - \dfrac{3}{4}x^3 + C$ , where $C$ is an arbitrary constant

$\dfrac{1}{4}x^2 - \dfrac{1}{2}x^3 + C$ , where $C$ is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q18:

2026: 22 May Shift 2

Calculus

>

Integrals

Easy

Common

Match List-I with List-II

List-I	List-II
(A) $\int_{-1}^{1} \lvert x \rvert dx =$	(I) 3
(B) $\int_{-\pi}^{\pi} \cos x \, dx =$	(II) 1
(C) $\int_{-1}^{1} (\lvert x \rvert-1) dx =$	(III) 0
(D) $\int_{-1}^{1} (\lvert x \rvert+1) dx =$	(IV) -1

Choose the correct answer from the options given below:

(A) - (I), (B) - (II), (C) - (IV), (D) - (III)

(A) - (II), (B) - (IV), (C) - (III), (D) - (I)

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

(A) - (I), (B) - (IV), (C) - (II), (D) - (III)

Correct Answer

Option 3

Correct Answer

Explanation

Q19:

2026: 22 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)

$\frac{1}{\log x}+C$

$\frac{x^2}{\log x}+C$

$\frac{x}{\log x}+C$

$\frac{1}{x\log x}+C$

Correct Answer

Option 3

Correct Answer

Explanation

Q20:

2026: 22 May Shift 2

Calculus

>

Integrals

Hard

Core

If $\int \frac{(\sqrt{x})^5}{(\sqrt{x})^7+x^6} dx = a\log\left|\frac{x^k}{1+x^k}\right| + C$ ; Where C is an arbitrary constant, then the value of $a.k$ is:

2/5

5/2

1

7/2

Correct Answer

Option 3

Correct Answer

Explanation

Q21:

2026: 22 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int \frac{2\cos 2x \, dx}{(\sin x+\cos x)^2}$ is equal to:(where C is an arbitrary constant)

$\frac{-4}{\sin x+\cos x}+C$

$2\log\lvert \sin x-\cos x \rvert+C$

$\log\lvert \sin x+\cos x \rvert+C$

$2\log\lvert \sin x+\cos x \rvert+C$

Correct Answer

Option 4

Correct Answer

Explanation

Q22:

2026: 22 May Shift 2

Calculus

>

Integrals

Easy

Core

$\int_0^{2\pi} \cos^5 x \, dx$ is equal to

1

-1

0

2

Correct Answer

Option 3

Correct Answer

Explanation

Q23:

2026: 22 May Shift 2

Calculus

>

Integrals

Hard

Applied

$\int_{1/3}^{1} \frac{(x-x^3)^{1/3}}{x^4} dx$ equals

$\frac{16}{3}$

8

6

$\frac{15}{2}$

Correct Answer

Option 3

Correct Answer

Explanation

Q24:

2026: 22 May Shift 2

Calculus

>

Integrals

Hard

Applied

Let $f(x)=\int \frac{3x-2}{(x+1)^2(x+3)}dx = \alpha\log\left|\frac{x+1}{x+3}\right|+\beta\left(\frac{1}{x+1}\right)+\gamma$ , where $\gamma$ is the constant of integration. If $f(0)=\frac{11}{4}\log\left(\frac{1}{3}\right)+4$ , then which of the following are correct?

(A) $4\alpha+2\beta+6\gamma=22$

(B) $f(1)=\alpha[1-\log 2]$

(C) $\alpha+\beta+\gamma=\frac{27}{4}$

(D) $\gamma=-\frac{5}{2}$

Choose the correct answer from the options given below:

(A), (B) and (C) only

(B), (C) and (D) only

(A) and (D) only

(B) and (C) only

Correct Answer

Option 4

Correct Answer

Explanation

Q25:

2026: 22 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int \frac{\sqrt{9+(\log x)^2}}{x} dx$ is equal to: (where $C$ is an arbitrary constant and consider $\log_e x = \log x$ )

$\frac{1}{2}x\sqrt{1+(\log x)^2} + 9\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\frac{1}{2}\log x\sqrt{9+(\log x)^2} + \frac{9}{2}\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\log x\sqrt{9+(\log x)^2} - \frac{9}{2}\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\log x\sqrt{9+(\log x)^2} - 9\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

Correct Answer

Option 2

Correct Answer

Explanation

Q26:

2026: 22 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx$ is equal to : (Consider $\log_e x = \log x$ )

1

2

3

4

Correct Answer

Option 3

Correct Answer

Explanation

Q27:

2026: 22 May Shift 1

Calculus

>

Integrals

Medium

Common

Match List-I with List-II

(Consider $\log_e x = \log x$ )

List-I	List-II
Integral	Solution: where C is an arbitrary constant
(A) $\displaystyle\int \dfrac{f'(x)}{f(x)\log f(x)}\,dx$	(I) $x\log(\log x) + c$
(B) $\displaystyle\int \left(\log(\log x) + \dfrac{1}{\log x}\right) dx$	(II) $\dfrac{x}{\log x} + C$
(C) $\displaystyle\int \dfrac{\log x}{(1+\log x)^2}\,dx$	(III) $\log\left\vert \log(f(x)) \right\vert + c$
(D) $\displaystyle\int \left(\dfrac{1}{\log x} - \dfrac{1}{(\log x)^2}\right) dx$	(IV) $\dfrac{x}{1+\log x} + c$

Choose the correct answer from the options given below:

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

(A) - (I), (B) - (II), (C) - (IV), (D) - (III)

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)

(A) - (III), (B) - (II), (C) - (IV), (D) - (I)

Correct Answer

Option 3

Correct Answer

Explanation

Q28:

2026: 22 May Shift 1

Calculus

>

Integrals

Easy

Core

$\int_{-1}^{1}\log(x+\sqrt{x^2+1})dx$ is equal to

$-\log2$

$\log2$

0

$\frac{1}{2}\log2$

Correct Answer

Option 3

Correct Answer

Explanation

Q29:

2026: 22 May Shift 1

Calculus

>

Integrals

Easy

Core

Value of $\int_{-2}^{2}|2x+1|dx$ is:

$\frac{17}{2}$

$\frac{17}{4}$

$\frac{9}{2}$

$\frac{9}{4}$

Correct Answer

Option 1

Correct Answer

Explanation

Q30:

2026: 22 May Shift 1

Calculus

>

Integrals

Easy

Core

Match List-I with List-II

Where C is arbitrary constant.

List-I	List-II
(A) $\displaystyle\int \dfrac{dx}{x^2 - a^2} =$	(I) $\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
(B) $\displaystyle\int \dfrac{dx}{a^2 - x^2} =$	(II) $\log\left\vert x + \sqrt{x^2-a^2}\right\vert + C$
(C) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}} =$	(III) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a}\right\vert + C$
(D) $\displaystyle\int \dfrac{dx}{\sqrt{x^2 + a^2}} =$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{a+x}{a-x}\right\vert + C$

Choose the correct answer from the options given below:

(A) - (III), (B) - (IV), (C) - (II), (D) - (I)

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

(A) - (II), (B) - (I), (C) - (IV), (D) - (III)

Correct Answer

Option 1

Correct Answer

Explanation

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