CUET Mathematics - Calculus - Integrals | Free Questions - Page 4

Q1:

2026: 18 May Shift 2

Calculus

>

Integrals

Hard

Core

If $f(x)=\int \sqrt{\dfrac{x+8}{x-8}}\,dx; f(8)=24\log_e 2$ , then $f(10)$ is equal to:

$6+32\log_e 2$

$6+\tan^{-1}\dfrac{4}{5}$

$-5+\sin^{-1}\dfrac{4}{5}$

$6+\dfrac{1}{8}\log_e\left|\dfrac{4}{5}\right|$

Correct Answer

Option 1

Correct Answer

Explanation

Q2:

2026: 18 May Shift 2

Calculus

>

Integrals

Easy

Applied

$\int \dfrac{e^{7\log x}-e^{6\log x}}{e^{5\log x}-e^{4\log x}}\,dx$ , $x>0$ and $x\ne 1$ is equal to (consider $\log x = \log_e x$ )

$\dfrac{x^5}{5}+C$ , where $C$ is an arbitrary constant.

$\dfrac{x^7}{7}+C$ , where $C$ is an arbitrary constant.

$\dfrac{x^8}{8}+C$ , where $C$ is an arbitrary constant.

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

Correct Answer

Option 4

Correct Answer

Explanation

Q3:

2026: 18 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int \frac{dx}{1+\sqrt{x}}$ is

$\sqrt{x}(1+\sqrt{x}) + C$ : where C is an arbitrary constant

$\sqrt{x} + \log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

$\sqrt{x} - \log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

$2\sqrt{x} - 2\log_e|1+\sqrt{x}| + C$ : where C is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q4:

2026: 18 May Shift 1

Calculus

>

Integrals

Easy

Common

Match List-I with List-II

List-I	List-II
Integral	Value of the integral
(A) $\displaystyle\int_{1}^{2} e^{2\log x} \, dx$	(I) $\log 2$
(B) $\displaystyle\int_{-1}^{1} \left\vert x \right\vert \, dx$	(II) $\dfrac{1}{2}$
(C) $\displaystyle\int_{0}^{1} \log e^x \, dx$	(III) $\dfrac{7}{3}$
(D) $\displaystyle\int_{1}^{2} \dfrac{1}{x} \, dx$	(IV) $1$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q5:

2026: 18 May Shift 1

Calculus

>

Integrals

Easy

Core

$\int \frac{1}{x^2+4x+8}dx$ is equal to (Where C is an arbitrary constant)

$-2\tan^{-1}(x+2)$

$\tan^{-1}\frac{x+2}{2} + C$

$-\frac{1}{2}\tan^{-1}\frac{x+2}{2} + C$

$\frac{1}{2}\tan^{-1}\frac{x+2}{2} + C$

Correct Answer

Option 4

Correct Answer

Explanation

Q6:

2026: 18 May Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 2

Correct Answer

Explanation

Q7:

2026: 18 May Shift 1

Calculus

>

Integrals

Easy

Core

If $\int \frac{1}{\tan x + \cot x}dx = k\cos 2x + C$ , where C is an arbitrary constant, then the value of $k$ is

1/2

-1/2

1/4

-1/4

Correct Answer

Option 4

Correct Answer

Explanation

Q8:

2026: 18 May Shift 1

Calculus

>

Integrals

Medium

Applied

If $\int e^{3\log x}(x^4+1)^{-1}dx = \frac{m}{n}\log(x^4+1)+C$ , where C is an arbitrary constant, $m$ and $n$ are positive integers such that g.c.d $(m,n)=1$ , then $(4m^2+n^2)$ is equal to

20

16

12

10

Correct Answer

Option 1

Correct Answer

Explanation

Q9:

2026: 15 May Shift 1

Calculus

>

Integrals

Medium

Common

Which of the following statements are TRUE?

A. $\displaystyle\int_4^9 \dfrac{1}{\sqrt{x}}\, dx = 2$

B. $\displaystyle\int_{-4}^4 (x^{101} + x^{201} + x^{301})\, dx = 4\left(\dfrac{1}{102} + \dfrac{1}{202} + \dfrac{1}{302}\right)$

C. $\displaystyle\int_{-1}^1 x^{1012}\, dx = \dfrac{2}{1013}$

D. $\displaystyle\int (x^2 + 2x + 1)\, dx = \dfrac{(x+1)^3}{3} + c$ where $c$ is an arbitrary constant

Choose the correct answer from the options given below:

A, C and D only

B, C and D only

B and C only

A, B, C and D

Correct Answer

Option 1

Correct Answer

Explanation

Q10:

2026: 15 May Shift 1

Calculus

>

Integrals

Medium

Common

$\displaystyle\int \dfrac{dx}{(x^2+4)^{\frac{3}{2}}}$ is equal to ( where $c$ is an arbitrary constant)

$\dfrac{x}{4\sqrt{x^2+4}} + c$

$\dfrac{4}{\sqrt{x^2+4}} + c$

$\dfrac{1}{4\sqrt{x^2+4}} + c$

$\dfrac{1}{\sqrt{x^2+4}} + c$

Correct Answer

Option 1

Correct Answer

Explanation

Q11:

2026: 15 May Shift 1

Calculus

>

Integrals

Easy

Core

If C is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\displaystyle\int \sqrt{x^2+a^2}\, dx$	I.	$\dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C$
B.	$\displaystyle\int \sqrt{a^2-x^2}\, dx$	II.	$\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$
C.	$\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\, dx$	III.	$\sin^{-1}\dfrac{x}{a} + C$
D.	$\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\, dx$	IV.	$\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left\vert x + \sqrt{x^2+a^2}\right\vert + C$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q12:

2026: 15 May Shift 1

Calculus

>

Integrals

Hard

Core

Value of $\displaystyle\int \dfrac{dx}{\sin^{\frac{3}{5}}x \cdot \cos^{\frac{7}{5}}x}$ is (where c is an arbitrary constant)

$\dfrac{5}{2}\tan^{\frac{2}{5}}x + c$

$\dfrac{2}{5}\tan^{\frac{5}{2}}x + c$

$\dfrac{3}{5}\sin^{\frac{4}{5}}x + c$

$\dfrac{1}{5}\cos^{\frac{1}{5}}x + c$

Correct Answer

Option 1

Correct Answer

Explanation

Q13:

2026: 15 May Shift 1

Calculus

>

Integrals

Hard

Core

If '[.]' is greatest integer function, then value of $\displaystyle\int_0^{1.5}[x^2]\,dx$ is equal to :

$2-\sqrt2$

$2+\sqrt2$

1.25

2.25

Correct Answer

Option 1

Correct Answer

Explanation

Q14:

2026: 15 May Shift 1

Calculus

>

Integrals

Easy

Applied

If $\displaystyle\int_0^3 (8x^3+5x^2-2x-k)\,dx = 3$ , where $k$ is constant, then $k$ is equal to

70

67

65

61

Correct Answer

Option 3

Correct Answer

Explanation

Q15:

2026: 14 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int \frac{x^8}{(x^3+1)^{1/3}} dx$ is equal to

$\frac{1}{4}(x^3+1)^{4/3} - \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{4}(x^3+1)^{4/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} - \frac{1}{2}(x^3+1)^{2/3} + \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q16:

2026: 14 May Shift 2

Calculus

>

Integrals

Easy

Common

If $\int_0^1 \frac{1-x}{1+x} dx = A\log 2 + B$ , then the value of A and B are

$A=2, B=2$

$A=2, B=1$

$A=2, B=-1$

$A=1, B=2$

Correct Answer

Option 3

Correct Answer

Explanation

Q17:

2026: 14 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int \frac{\sqrt{1+x^2}}{x^4} dx$ is equal to :

$\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :Where C is an arbitrary constant

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{5/2}+C$ :Where C is an arbitrary constant

Correct Answer

Option 2

Correct Answer

Explanation

Q18:

2026: 14 May Shift 2

Calculus

>

Integrals

Medium

Core

Match the LIST-I with LIST-II

	LIST-I (Definite integrals)		LIST-II (Value)
A.	$\displaystyle\int_{-6}^{6} \vert x \vert \, dx$	I.	$1$
B.	$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^{2017} x \, dx$	II.	$36$
C.	$\displaystyle\int_{0}^{\pi/4} \cos x \, dx$	III.	$0$
D.	$\displaystyle\int_{0}^{2} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}} \, dx$	IV.	$\dfrac{1}{\sqrt{2}}$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q19:

2026: 14 May Shift 2

Calculus

>

Integrals

Easy

Core

If C is an arbitrary constant, then

Match the LIST-I with LIST-II

	LIST-I		LIST-II
A.	$\displaystyle\int a^x \, dx$	I.	$\sin^{-1}\dfrac{x}{a} + C$
B.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 - x^2}}$	II.	$\log_e \vert x + \sqrt{x^2 - a^2} \vert + C$
C.	$\displaystyle\int \dfrac{dx}{\sqrt{a^2 + x^2}}$	III.	$\dfrac{a^x}{\log_e a} + C$
D.	$\displaystyle\int \dfrac{dx}{\sqrt{x^2 - a^2}}$	IV.	$\log_e \vert x + \sqrt{a^2 + x^2} \vert + C$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q20:

2026: 12 May Shift 2

Calculus

>

Integrals

Medium

Common

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

$x\log_e x + C$

$\dfrac{x}{\log_e x}+C$

$e^x\log_e x + C$

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

Correct Answer

Option 2

Correct Answer

Explanation

Q21:

2026: 12 May Shift 2

Calculus

>

Integrals

Easy

Common

If $\displaystyle\int_0^{40} \dfrac{dx}{2x+1} = \log_e k$ , then the value of $k$ is :

3

$\dfrac{9}{2}$

9

$\dfrac{3}{2}$

Correct Answer

Option 3

Correct Answer

Explanation

Q22:

2026: 12 May Shift 2

Calculus

>

Integrals

Easy

Core

Match List-I with List-II

List-I	List-II
Indefinite integral	Solution (where 'c' is an arbitrary constant)
(A) $\displaystyle\int \sqrt{x^2-a^2} \, dx$	(I) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c$
(B) $\displaystyle\int \dfrac{1}{x^2-a^2} \, dx$	(II) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c$
(C) $\displaystyle\int \dfrac{1}{x^2+a^2} \, dx$	(III) $\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c$
(D) $\displaystyle\int \sqrt{x^2+a^2} \, dx$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 3

Correct Answer

Explanation

Q23:

2026: 12 May Shift 2

Calculus

>

Integrals

Medium

Core

If $\displaystyle\int\dfrac{x^3\,dx}{\sqrt{1+x^2}} = a(1+x^2)^{3/2}+b\sqrt{1+x^2}+c$ , where 'C' is an arbitrary constant, then the value of $a$ and $b$ are

$a=\dfrac{1}{3}, b=1$

$a=-\dfrac{1}{3}, b=1$

$a=-\dfrac{1}{3}, b=-1$

$a=\dfrac{1}{3}, b=-1$

Correct Answer

Option 4

Correct Answer

Explanation

Q24:

2026: 12 May Shift 2

Calculus

>

Integrals

Medium

Core

The value of $\displaystyle\int_{-1}^{1}|x^4-x|\,dx$ is equal to:

$\dfrac{1}{2}$

$\dfrac{1}{3}$

2

1

Correct Answer

Option 4

Correct Answer

Explanation

Q25:

2026: 12 May Shift 2

Calculus

>

Integrals

Easy

Applied

Match the rational functions with the correct form of the partial fractions.

[ $p,q,r,a,b,c,A,B,C$ are arbitrary constants]

Match List-I with List-II

List-I (Rational Function)	List-II (Partial Fraction)
(A) $\dfrac{px^2+qx+r}{(x-a)(x^2+bx+c)}$	(I) $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$
(B) $\dfrac{px^2+qx+r}{(x-a)^2(x-b)}, a\neq b$	(II) $\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$
(C) $\dfrac{px^2+qx+r}{(x-a)(x-b)(x-c)}, a\neq b\neq c$	(III) $\dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+bx+c}$
(D) $\dfrac{px+q}{(x-a)^2}$	(IV) $\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q26:

2026: 12 May Shift 2

Calculus

>

Integrals

Medium

Applied

The value of the integral $I=\displaystyle\int\dfrac{e^x(x-2)}{(x+1)^4}dx$ is, [ $C$ is an arbitrary constant]

$\dfrac{e^x}{(x+1)^4}+C$

$\dfrac{e^x}{(x+1)^3}+C$

$\dfrac{-3e^x}{(x+1)^4}+C$

$\dfrac{3e^x}{(x+1)^3}+C$

Correct Answer

Option 2

Correct Answer

Explanation

Q27:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Common

$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$ is equal to

$a$

$\frac{a}{2}$

$\frac{a}{3}$

$2a$

Correct Answer

Option 2

Correct Answer

Explanation

Q28:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Common

$\int_0^8 (x^{\frac{2}{3}} + 1) dx$ is equal to

$\frac{28}{3}$

$\frac{25}{3}$

$\frac{184}{3}$

$\frac{136}{5}$

Correct Answer

Option 4

Correct Answer

Explanation

Q29:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Core

$\int e^{2x}(\sin x + \frac{1}{2}\cos x) dx$ is equal to

$-\frac{1}{2}e^{2x}\cos x + C$ , C is an arbitrary constant

$\frac{1}{2}e^{2x}\sin x + C$ , C is an arbitrary constant

$e^{2x}\sin x + C$ , C is an arbitrary constant

$-e^{2x}\cos x + C$ , C is an arbitrary constant

Correct Answer

Option 2

Correct Answer

Explanation

Q30:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Core

If $\int \sqrt\frac{1-x}{{1+x}} dx = a\sqrt{1-x^2} + \beta \sin^{-1}x + C$ , Where C is an arbitrary constant, then which of the following are TRUE?

(A) $\alpha = 1$

(B) $\alpha = -1$

(C) $\beta = 1$

(D) $\beta = -1$

Choose the correct answer from the options given below:

(B) and (D) only

(A) and (C) only

(A) and (D) only

(B) and (C) only

Correct Answer

Option 2

Correct Answer

Explanation

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