CUET Mathematics - Calculus - Integrals | Free Questions - Page 6

Q1:

2025: 29 May Shift 2

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
Integral	Solution: C is an arbitrary constant
---	---
(A) $\int \frac{dx}{x^2 + 25}$	(I) $\frac{1}{10} \log \left\vert \frac{5 + x}{5 - x} \right\vert + C$
(B) $\int \frac{dx}{x^2 - 25}$	(II) $\log \vert x + \sqrt{x^2 - 25}\vert + C$
(C) $\int \frac{dx}{25 - x^2}$	(III) $\frac{1}{5} \tan^{-1} \left( \frac{x}{5} \right) + C$
(D) $\int \frac{dx}{\sqrt{x^2 - 25}}$	(IV) $\frac{1}{10} \log \left\vert \frac{x - 5}{x + 5} \right\vert + C$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 2

Correct Answer

Explanation

Q2:

2025: 29 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{(1+5\sin^2 x)}$ is equal to

$\sqrt{6}\cot^{-1}(\sqrt{6}\cot x) + C$ ; C is an arbitrary constant

$\sqrt{6}\cot^{-1}(\sqrt{6} \tan x) + C$ ; C an arbitrary constant

$\frac{1}{\sqrt{6}} \tan^{-1}(\sqrt{6}\cot x) + C$ ; C an arbitrary constant

$\frac{1}{\sqrt{6}} \tan^{-1}(\sqrt{6} \tan x) + C$ ; C an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q3:

2025: 29 May Shift 2

Calculus

>

Integrals

Easy

Core

If $\int_0^a 3x^2dx = 8$ , then the value of $a$ is:

2

10

8

5

Correct Answer

Option 1

Correct Answer

Explanation

Q4:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int \frac{x^5}{\sqrt{1 + x^3}} dx$ is

$\frac{2}{9}(1 + x^3)^{3/2} - \frac{2}{3}(1 + x^3)^{1/2} + c$ : $c$ is an arbitrary constant

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

$\frac{1}{3}(1 + x^3)^{3/2} + \frac{1}{3}(1 + x^3)^{1/2} + c$ : $c$ is an arbitrary constant

$\frac{2}{9}(1 + x^3)^{3/2} + \frac{2}{3}(1 + x^3)^{1/2} + c$ : $c$ is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q5:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int_{-1}^{1} |x^3 - x| dx$ is

0

$\frac{1}{4}$

$\frac{1}{2}$

1

Correct Answer

Option 3

Correct Answer

Explanation

Q6:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Core

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

$\log|e^x + e^{-x}| + C$ : $C$ is an arbitrary constant

$\log|e^{2x} + 1| + C$ : $C$ is an arbitrary constant

$\log|e^{2x} - e^{-x}| + C$ : $C$ is an arbitrary constant

$\log|e^{2x} - 1| + C$ : $C$ is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q7:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int e^{-x}(\cot x + \cosec^2 x)dx =$

$e^{-x}\cot x + c$ , where $c$ is an arbitrary constant.

$-e^{-x}\cot x + c$ , where $c$ is an arbitrary constant.

$-e^{-x}\cosec^2 x + c$ , where $c$ is an arbitrary constant.

$e^{-x}\cosec^2 x + c$ , where $c$ is an arbitrary constant.

Correct Answer

Option 2

Correct Answer

Explanation

Q8:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{e^x + e^{-x}}$ is equal to

$\tan^{-1}(e^x) + c$ : $c$ is an arbitrary constant

$\tan^{-1}(e^{-x}) + c$ : $c$ is an arbitrary constant

$log(e^x - e^{-x}) + c$ : $c$ is an arbitrary constant

$log(e^x + e^{-x}) + c$ : $c$ is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q9:

2025: 27 May Shift 1

Calculus

>

Integrals

Medium

Applied

The value of $\int_{-5}^{5} |x + 3| dx$ is

32

36

34

50

Correct Answer

Option 3

Correct Answer

Explanation

Q10:

2025: 26 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int_{1}^{2} \frac{1}{x(x+1)} dx, x > 0$ equals

$\log\left(\frac{2}{3}\right)$

$\log\left(\frac{4}{3}\right)$

0

$\log\left(\frac{3}{4}\right)$

Correct Answer

Option 2

Correct Answer

Explanation

Q11:

2025: 26 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)

$xe^x + C$

$\frac{e^x}{x} + C$

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

$e^x + \frac{1}{x} + C$

Correct Answer

Option 2

Correct Answer

Explanation

Q12:

2025: 26 May Shift 2

Calculus

>

Integrals

Hard

Core

$\int e^x \left(\frac{1-x}{1+x^2}\right)^2 dx$ is equal to

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

$\frac{e^x}{(1+x^2)^2} + C$ ; $C$ is an arbitrary constant

$\frac{-e^x}{(1+x^2)^2} + C$ ; $C$ is an arbitrary constant

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

Correct Answer

Option 1

Correct Answer

Explanation

Q13:

2025: 26 May Shift 2

Calculus

>

Integrals

Medium

Core

If $f(x)$ and $g(x)$ are continuous functions in [0, a] such that $f(x) = f(a - x)$ and $g(x) + g(a - x) = a$ then $\int_{0}^{a} f(x)g(x)dx =$

$a\int_{0}^{a} f(x)dx$

$a\int_{0}^{a} g(x)dx$

$\frac{a}{2}\int_{0}^{a} f(x)dx$

$\frac{a}{2}\int_{0}^{a} g(x)dx$

Correct Answer

Option 3

Correct Answer

Explanation

Q14:

2025: 26 May Shift 2

Calculus

>

Integrals

Medium

Core

The value of $\int_{0}^{\pi/2} \frac{\tan^7 x}{\cot^7 x + \tan^7 x} dx$ is

$\pi/2$

$\pi/3$

$\pi/4$

$\pi/6$

Correct Answer

Option 3

Correct Answer

Explanation

Q15:

2025: 26 May Shift 2

Calculus

>

Integrals

Medium

Applied

The integral $\int \frac{2dx}{e^{2x}-1}$ is equal to:

$\log|e^{2x} - 1| + C$ ; $C$ is an arbitrary constant

$\log\left|\frac{e^{2x}-1}{e^{2x}+1}\right| + C$ ; $C$ is an arbitrary constant

$\log\left|\frac{e^{2x}}{e^{2x}-1}\right| + C$ ; $C$ is an arbitrary constant

$\log\left|\frac{e^{2x}-1}{e^{2x}}\right| + C$ ; $C$ is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q16:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$ equals

$\sin^{-1}\left(\frac{\cos^2 x}{3}\right) + C$ ; $C$ is an arbitrary constant.

$\cos^{-1}\left(\frac{\sin^2 x}{3}\right) + C$ ; $C$ is an arbitrary constant

$\sin\left(\frac{\cos^2 x}{3}\right) + C$ ; $C$ is an arbitrary constant

$-\sin^{-1}\left(\frac{\cos^2 x}{3}\right) + C$ ; $C$ is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q17:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int_0^2 (|x| + |x - 2|) dx =$

8

4

2

$\frac{1}{2}$

Correct Answer

Option 2

Correct Answer

Explanation

Q18:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Core

If $\int_0^1 \frac{e^x}{1 + x} dx = m$ , then the value of $\int_0^1 \frac{e^x}{(1 + x)^2} dx$ is:

$m - e + 1$

$\frac{m}{2} + e - 1$

$m - \frac{e}{2} + 1$

$m + \frac{e}{2} - 1$

Correct Answer

Option 3

Correct Answer

Explanation

Q19:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{x^3\sqrt{(1 + x^4)}} =$

$\frac{1}{4x^2}\sqrt{1 + x^4} + C$ , $C$ is an arbitrary constant

$\frac{1}{x^2}\sqrt{1 + x^4} + C$ , $C$ is an arbitrary constant

$\frac{1}{2x^2}\sqrt{1 + x^4} + C$ , $C$ is an arbitrary constant

$-\frac{1}{2x^2}\sqrt{1 + x^4} + C$ , $C$ is an arbitary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q20:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Core

Match List-I with List-II (where $c$ is an arbitrary constant)

List-I	List-II
(A) $\int \tan x \, dx$	(I) $\log\vert \sec x + \tan x\vert + c$
(B) $\int \cot x \, dx$	(II) $\log\vert \sec x\vert + c$
(C) $\int \sec x \, dx$	(III) $\log\vert \sin x\vert + c$
(D) $\int \cosec x \, dx$	(IV) $\log\vert \cosec x - \cot x\vert + c$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 3

Correct Answer

Explanation

Q21:

2025: 22 May Shift 2

Calculus

>

Integrals

Medium

Applied

If $\int \frac{2x - 5}{(2x - 3)^3} e^{2x} dx = \frac{\lambda e^{2x}}{(2x - 3)^2} + C$ , where $C$ is an arbitrary constant then the value of $\lambda$ is

$\frac{1}{2}$

2

-2

$-\frac{1}{2}$

Correct Answer

Option 1

Correct Answer

Explanation

Q22:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int \frac{x}{(x-1)(x-2)} dx$ is equal to ( where $C$ is a constant of integration)

$\log_e\left|\frac{(x-1)^2}{x-2}\right| + C$

$\log_e\left|\frac{(x-2)^2}{x-1}\right| + C$

$\log_e|(x-2)(x-1)^2| + C$

$\log_e\left|\frac{x-1}{(x-2)^2}\right| + C$

Correct Answer

Option 2

Correct Answer

Explanation

Q23:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int_0^1 x e^x dx$ is equal to

0

1

e

$e^2$

Correct Answer

Option 2

Correct Answer

Explanation

Q24:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{2\sin^2 x + 5\cos^2 x}$ is equal to

$\frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{ \tan x}{\sqrt{5}}\right) + C$ ; C is an arbitrary constant

$\frac{1}{\sqrt{5}} \tan^{-1}\left(\frac{2 \tan x}{\sqrt{5}}\right) + C$ ; C is an arbitrary constant

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\sqrt{2} \tan x}{\sqrt{5}}\right) + C$ ; C is an arbitrary constant

$\frac{1}{\sqrt{10}} \tan^{-1}\left(\frac{\sqrt{2} \tan x}{\sqrt{5}}\right) + C$ ; C is an arbitrary constant

Correct Answer

Option 4

Correct Answer

Explanation

Q25:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Core

If $f(a-x) = f(x)$ , then $\int_0^a xf(x)dx$ is equal to

$\frac{a}{2}\int_0^a f(x)dx$

$2\int_0^a f(x)dx$

$a f(a)$

$a\int_0^a f(x)dx$

Correct Answer

Option 1

Correct Answer

Explanation

Q26:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Core

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

$\frac{x}{\log_e x} + c$ ; c is an arbitrary constant

$\frac{1}{\log_e x} + c$ ; c is an arbitrary constant

$\frac{x}{(\log_e x)^2} + c$ ; c is an arbitrary constant

$\frac{\log_e x}{x} + c$ ; c is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q27:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Applied

The Value of $\int_1^3 |2x - 1|dx$ equal to

2

4

6

8

Correct Answer

Option 3

Correct Answer

Explanation

Q28:

2025: 22 May Shift 1

Calculus

>

Integrals

Medium

Applied

The integral $\int e^x\left(\frac{x-1}{2x^2}\right)dx$ is equal to

$\frac{e^x}{2x} + C$ , where $C$ is constant of integration.

$\frac{2e^x}{x} + C$ , where $C$ is constant of integration.

$\frac{e^x}{x} + C$ , where $C$ is constant of integration.

$\frac{e^x}{4x} + C$ , where $C$ is constant of integration.

Correct Answer

Option 1

Correct Answer

Explanation

Q29:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Common

Value of $\int \left(\frac{1}{logx} - \frac{1}{(logx)^2}\right)dx$ is

$\frac{x}{logx} + c$ , where c is an arbitrary constant

$x\,logx + c$ , where c is an arbitrary constant

$\frac{1}{logx} + c$ , where c is an arbitrary constant

$\frac{e^x}{logx} + c$ , where c is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q30:

2025: 21 May Shift 2

Calculus

>

Integrals

Medium

Common

The value of $\int_1^3 \frac{x^2}{x^3+1}dx$

$\frac{1}{3}\log 26$

$\log 14$

$\frac{1}{3}\log 14$

$\log 56$

Correct Answer

Option 3

Correct Answer

Explanation

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