CUET Mathematics - Calculus - Integrals | Free Questions - Page 8

Q1:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Common

For $x \neq -1$ , if $\int \frac{xe^x dx}{(1+x)^2} = \frac{ae^x}{(1+x)^b} + c$ , where a, b are fixed numbers and c is the integration constant, then $a + b$ is equal to

0

1

2

3

Correct Answer

Option 3

Correct Answer

Explanation

Q2:

2025: 15 May Shift 1

Calculus

>

Integrals

Hard

Core

For $x \in \left(0, \frac{\pi}{2}\right)$ , $\int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx$ is equal to

$\sin^{-1}(\sin 2x) + C$ , C is an arbitary constant

$\sin^{-1}(\cos x - \sin x) + C$ , C is an arbitary constant

$\sin^{-1}(\sin x - \cos x) + C$ , C is an arbitary constant

$\sin^{-1}(\sin x + \cos x) + C$ , C is an arbitary constant

Correct Answer

Option 3

Correct Answer

Explanation

Q3:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Core

If $\int e^x\left(\frac{x-1}{(x+1)^3}\right)dx = \frac{Ae^x}{(x+1)^B} + C$ , where C is constant of integration then which of the following are correct?

(A) $A = -1$

(B) $A = 1$

(C) $B = 3$

(D) $B = 2$

Choose the correct answer from the options given below:

(A) and (C) only

(B) only

(B) and (D) only

(A) and (D) only

Correct Answer

Option 3

Correct Answer

Explanation

Q4:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Core

For $x \in \left(0, \frac{\pi}{2}\right)$ , $\int \frac{1}{\sin^2 x + \sin 2x} dx$ is equal to

$\frac{1}{2}\log\left|\frac{ \tan x + 2}{ \tan x - 2}\right| + C$ , where C is constant of integration

$\frac{1}{2}\log\left|\frac{\tan x}{ \tan x + 2}\right| + C$ , where C is constant of integration

$\frac{1}{2}\log\left|\frac{\tan x + 2}{ \tan x}\right| + C$ , where C is constant of integration

$\frac{1}{2}\log\left|\frac{\tan x + 1}{\tan x + 2}\right| + C$ , where C is constant of integration

Correct Answer

Option 2

Correct Answer

Explanation

Q5:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Applied

If $I = \int \frac{x^4 + x^2 + 1}{x^2 - x + 1} dx = \alpha x + \beta x^2 + \gamma x^3 + \delta$ , $\delta$ is constant of integration, then $(\alpha + 2\beta + 3\gamma)$ equals

0

1

2

3

Correct Answer

Option 4

Correct Answer

Explanation

Q6:

2025: 15 May Shift 1

Calculus

>

Integrals

Medium

Applied

The value of the integral $I = \int_0^1 \frac{1}{\sqrt{x^2 + 2x + 3}} dx$ is:

$\log\left(\frac{2 + \sqrt{6}}{1 - \sqrt{3}}\right)$

$\log\left(\frac{2 + \sqrt{6}}{1 + \sqrt{3}}\right)$

$\log\left(\frac{5 + \sqrt{6}}{1 + \sqrt{3}}\right)$

$\log(5 + \sqrt{6})$

Correct Answer

Option 2

Correct Answer

Explanation

Q7:

2025: 14 May Shift 2

Calculus

>

Integrals

Medium

Common

For $x \in \mathbb{R} - \{-1,0,1\}$ , $\int \frac{1}{x - x^5}dx$ is equal to

$\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right| + c$ : where $c$ is constant of integration.

$-\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right| + c$ : where $c$ is constant of integration.

$4\log_e\left|\frac{x^4}{1-x^4}\right| + c$ : where $c$ is constant of integration.

$-4\log_e\left|\frac{x^4}{1-x^4}\right| + c$ : where $c$ is constant of integration.

Correct Answer

Option 1

Correct Answer

Explanation

Q8:

2025: 14 May Shift 2

Calculus

>

Integrals

Easy

Common

$\int_{-1}^{1}(x^7 + x^5 + x^3 + x + 1)dx$ is equal to

0

1

2

4

Correct Answer

Option 3

Correct Answer

Explanation

Q9:

2025: 14 May Shift 2

Calculus

>

Integrals

Hard

Core

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

$\tan(e^x) + c$ , where $c$ is constant of integration.

$-\cot(ex^x) + c$ , where $c$ is constant of integration.

$\cot(e^x) + c$ , where $c$ is constant of integration.

$\tan(xe^x) + c$ , where $c$ is constant of integration.

Correct Answer

Option 4

Correct Answer

Explanation

Q10:

2025: 14 May Shift 2

Calculus

>

Integrals

Medium

Core

$\int_{-\pi}^{\pi} \frac{e^{\sin x}}{e^{\sin x} + e^{-\sin x}}dx$ is equal to

0

$\frac{\pi}{2}$

$\pi$

$\frac{\pi}{4}$

Correct Answer

Option 3

Correct Answer

Explanation

Q11:

2025: 14 May Shift 2

Calculus

>

Integrals

Hard

Core

If $I_n = \int_{0}^{\pi/4} \tan^n x dx$ then $I_{2024} + I_{2026}$ is equal to:

$\frac{1}{2025}$

$\frac{1}{2027}$

$\frac{1}{2023}$

$\frac{2}{2025}$

Correct Answer

Option 1

Correct Answer

Explanation

Q12:

2025: 14 May Shift 2

Calculus

>

Integrals

Hard

Applied

If $\int \frac{dx}{(x-1)^3/^4. (x+2)^5/^4} = \alpha[1 - g(x)]^\beta + c$ , where $c$ is a constant of integration, then which of the following are true?

(A) $\alpha = \frac{2}{3}$

(B) $\beta = \frac{3}{4}$

(C) $3\alpha + 4\beta = 5$

(D) $g(x) = \frac{3}{(x+2)}$

Choose the correct answer from the options given below:

(A), (B) and (D) only

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

(A), (B), (C) and (D)

(C) and (D) only

Correct Answer

Option 4

Correct Answer

Explanation

Q13:

2025: 14 May Shift 2

Calculus

>

Integrals

Medium

Applied

The value of the definite integral $I = \int_{1}^{2} \frac{1}{x(1 + x^2)}dx$ is:

3/2

$\frac{3}{2}\log 2 - \frac{1}{2}\log 5$

$\frac{5}{2}\log 3 - \frac{1}{2}\log 5$

$\frac{5}{2}\log 3 + \log 5$

Correct Answer

Option 2

Correct Answer

Explanation

Q14:

2025: 14 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int_1^4 |x - 2|dx$ is equal to

5

$\frac{7}{2}$

$\frac{3}{2}$

$\frac{5}{2}$

Correct Answer

Option 4

Correct Answer

Explanation

Q15:

2025: 14 May Shift 1

Calculus

>

Integrals

Medium

Common

The integral I = $\int \frac{e^{5\log_e x} - e^{4\log_e x}}{e^{3\log_e x} - e^{2\log_e x}} dx$ is equal to

$\frac{x}{2} + C$ , where C is the constant of integration

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

$\frac{x^3}{3} + C$ , where C is the constant of integration

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

Correct Answer

Option 3

Correct Answer

Explanation

Q16:

2025: 14 May Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
Definite integral	Value
(A) $\int_0^1 \frac{2x}{1 + x^2} dx$	(I) 2
(B) $\int_{-1}^1 sin^3 x \cos^4 x dx$	(II) $log_e\left(\frac{3}{2}\right)$
(C) $\int_0^{\pi} \sin x dx$	(III) $log_e 2$
(D) $\int_2^3 \frac{2}{x^2 - 1} dx$	(IV) 0

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q17:

2025: 14 May Shift 1

Calculus

>

Integrals

Medium

Core

The integral I = $\int e^x\left(\frac{x - 1}{3x^2}\right) dx$ is equal to

$\frac{1}{3}\left(\frac{x^2}{2} - x\right) + C$ , where C is constant of integration

$\left(\frac{x^2}{2} - x\right)e^x + C$ , where C is constant of integration

$\frac{1}{3x^2}e^x + C$ , where C is constant of integration

$\frac{1}{3x}e^x + C$ , where C is constant of integration

Correct Answer

Option 4

Correct Answer

Explanation

Q18:

2025: 14 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int_{\pi/6}^{\pi/3} \frac{tan x}{tan x + cot x} dx$ is equal to

$\frac{\pi}{4}$

0

$\frac{\pi}{6}$

$\frac{\pi}{12}$

Correct Answer

Option 4

Correct Answer

Explanation

Q19:

2025: 14 May Shift 1

Calculus

>

Integrals

Hard

Applied

If $\int \frac{(1 + x \log x)}{xe^{-x}} dx = e^x f(x) + C$ , where C is constant of integration, then $f(x)$ is

$e^x$

$\log x$

$\frac{1}{x}$

$\frac{1}{x^2}$

Correct Answer

Option 2

Correct Answer

Explanation

Q20:

2025: 13 May Shift 2

Calculus

>

Integrals

Medium

Common

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

$\frac{1}{log_e t} + C$ , where C is constant of integration

$\frac{t}{log_e t} + C$ , where C is constant of integration

$\frac{-t}{log_e t} + C$ , where C is constant of integration

$\frac{-1}{log_e t} + C$ , where C is constant of integration

Correct Answer

Option 2

Correct Answer

Explanation

Q21:

2025: 13 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int_0^2 x(2-x)^n dx$ is equal to

$\frac{2^{n+2}(n+1)}{(n+2)(n+3)}$

$\frac{2^{n+2}(n+2)}{(n+1)(n+3)}$

$\frac{2^{n+2}}{(n+1)(n+2)}$

$\frac{2^{n+1}}{(n+1)(n+2)}$

Correct Answer

Option 3

Correct Answer

Explanation

Q22:

2025: 13 May Shift 2

Calculus

>

Integrals

Hard

Core

$\int_0^1 tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ is equal to

-1

0

1

$\frac{\pi}{4}$

Correct Answer

Option 2

Correct Answer

Explanation

Q23:

2025: 13 May Shift 2

Calculus

>

Integrals

Easy

Core

$\int_{-\frac{5}{2}}^{\frac{5}{2}} |x| dx$ is equal to

$\frac{25}{4}$

0

$\frac{5}{2}$

$-\frac{5}{2}$

Correct Answer

Option 1

Correct Answer

Explanation

Q24:

2025: 13 May Shift 2

Calculus

>

Integrals

Medium

Core

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

$\frac{1}{\cos x + \sin x} + C$ , where C is constant of integration

$\frac{-1}{\cos x + \sin x} + C$ , where C is constant of integration

$\frac{1}{1 + \sin 2x} + C$ , where C is constant of integration

$\frac{1}{1 - \sin 2x} + C$ , where C is constant of integration

Correct Answer

Option 2

Correct Answer

Explanation

Q25:

2025: 13 May Shift 2

Calculus

>

Integrals

Medium

Applied

If the integral $I = \int \left[log_e(log_e x)^2 + \frac{a}{log_e x}\right] dx = x log_e(log_e x)^2 + C$ , where C is constant of integration. Then the value of $a$ is:

0

-1

-2

2

Correct Answer

Option 4

Correct Answer

Explanation

Q26:

2025: 13 May Shift 1

Calculus

>

Integrals

Easy

Common

$\int \frac{x^3 - 1}{x^2} dx$ is equal to

$\frac{x^2}{2} + x + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} - \frac{1}{x} + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} - x + c$ , Where $c$ is constant of integration

$\frac{x^2}{2} + \frac{1}{x} + c$ , Where $c$ is constant of integration

Correct Answer

Option 4

Correct Answer

Explanation

Q27:

2025: 13 May Shift 1

Calculus

>

Integrals

Medium

Common

The value of which of the following integrals is zero?

(A) $\int_0^1 x dx$

(B) $\int_{-1}^1 x dx$

(C) $\int_{-1}^1 x^2 dx$

(D) $\int_0^1 \log\left(\frac{x}{1-x}\right) dx$

Choose the correct answer from the options given below:

(B) and (D) only

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

(B) only

(C) and (D) only

Correct Answer

Option 1

Correct Answer

Explanation

Q28:

2025: 13 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{9x^2 - 16}$ is equal to

$\frac{1}{24}\log_e\left|\frac{3x+4}{3x-4}\right| + C$ , Where $C$ is constant of integration

$\frac{3}{8}\log_e\left|\frac{3x+4}{3x-4}\right| + C$ , Where $C$ is constant of integration

$\frac{3}{8}\log_e\left|\frac{3x-4}{3x+4}\right| + C$ , Where $C$ is constant of integration

$\frac{1}{24}\log_e\left|\frac{3x-4}{3x+4}\right| + C$ , Where $C$ is constant of integration

Correct Answer

Option 4

Correct Answer

Explanation

Q29:

2025: 13 May Shift 1

Calculus

>

Integrals

Hard

Core

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{tanx}}$ is equal to

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{\pi}{6}$

$\frac{\pi}{12}$

Correct Answer

Option 4

Correct Answer

Explanation

Q30:

2025: 13 May Shift 1

Calculus

>

Integrals

Hard

Core

$\int \sin x \sin 2x \sin 3x dx$ is equal to

$-\frac{1}{48}(6\cos 2x + 3\cos 4x + 2\cos 6x) + C$ , Where $C$ is constant of integration

$-\frac{1}{48}(2\cos 2x + 3\cos 4x - 6\cos 6x) + C$ , Where $C$ is constant of integration

$-\frac{1}{48}(6\cos 2x + 3\cos 4x - 2\cos 6x) + C$ , Where $C$ is constant of integration

$-\frac{1}{48}(2\cos 2x + 3\cos 4x + 6\cos 6x) + C$ , Where $C$ is constant of integration

Correct Answer

Option 3

Correct Answer

Explanation

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