CUET Mathematics - Calculus - Integrals | Free Questions - Page 5

Q1:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
(A) $\int \dfrac{dx}{x^2 - 16}$	(I) $\dfrac{1}{8} \log \left\vert \dfrac{4 + x}{4 - x} \right\vert + c$ , Where C is an arbitrary constant,
(B) $\int \dfrac{dx}{x^2 + 16}$	(II) $\log \left\vert x + \sqrt{x^2 - 16} \right\vert + c$ , Where C is an arbitrary constant,
(C) $\int \dfrac{dx}{16 - x^2}$	(III) $\dfrac{1}{8} \log \left\vert \dfrac{x - 4}{x + 4} \right\vert + c$ , Where C is an arbitrary constant,
(D) $\int \dfrac{dx}{\sqrt{x^2 - 16}}$	(IV) $\dfrac{1}{4} \tan^{-1} \left( \dfrac{x}{4} \right) + c$ , Where C is an arbitrary constant,

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q2:

2025: 3 June Shift 2

Calculus

>

Integrals

Medium

Applied

The value of $\int_2^4 \frac{x}{x^2 + 1} dx$ is equal to

$\frac{1}{2}\log_e(\frac{17}{5})$

$\frac{1}{5}\log_e(\frac{17}{5})$

$\log_e(\frac{17}{5})$

$2\log_e(\frac{17}{5})$

Correct Answer

Option 1

Correct Answer

Explanation

Q3:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Common

$\int \sqrt{1 + \frac{x^2}{9}} dx$ is equal to

(Where C is an arbitrary constant)

$\frac{1}{9}[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}] + C$

$3[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}] + C$

$[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}] + C$

$\frac{1}{3}[\frac{9}{2}\log|x + \sqrt{x^2 + 9}| + \frac{x}{2}\sqrt{x^2 + 9}] + C$

Correct Answer

Option 4

Correct Answer

Explanation

Q4:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Common

$\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$ is equal to

$\frac{3}{2}$

$\frac{1}{2}$

1

0

Correct Answer

Option 2

Correct Answer

Explanation

Q5:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Core

$\int \frac{dx}{\sqrt{5 - 4x - x^2}}$ is equal to

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

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

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

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

Correct Answer

Option 1

Correct Answer

Explanation

Q6:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Core

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

$\frac{1}{(1 + \log_e x)^2} + C$ , where C is constant of integration

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

$\frac{x}{(1 + \log_e x)^2} + C$ , where C is constant of integration

$\frac{1}{1 + \log_e x} + C$ , whereC is constant of integration

Correct Answer

Option 2

Correct Answer

Explanation

Q7:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
Integral	Value
---	---
(A) $\int_{-1}^{1} (\vert x\vert + 1) dx$	(I) 0
(B) $\int_{-2}^{2} \vert x + 1\vert dx$	(II) 2
(C) $\int_{-1}^{1} 3\vert x^2\vert dx$	(III) 5
(D) $\int_{-1}^{1} x\vert x\vert dx$	(IV) 3

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 4

Correct Answer

Explanation

Q8:

2025: 3 June Shift 1

Calculus

>

Integrals

Medium

Applied

The value of $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to (where C is an arbitrary constant)

$\frac{4}{15}(1 - \frac{1}{x^3})^{5/4} + C$

$\frac{4}{15}(1 + \frac{1}{x^3})^{5/4} + C$

$\frac{4}{15}(1 - \frac{1}{x^3})^{4/5} + C$

$\frac{4}{15}(1 + \frac{1}{x^3})^{4/5} + C$

Correct Answer

Option 1

Correct Answer

Explanation

Q9:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Common

The value of $\int_0^1 \log_e\left(\frac{1}{x} - 1\right)dx$ is:

1

0

3/2

$\log 2$

Correct Answer

Option 2

Correct Answer

Explanation

Q10:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Common

$\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$ is equal to

$\frac{1}{2}\log_e|e^{2x} - e^{-2x}| + C$ , where $C$ is an arbitrary constant.

$\log_e|e^{2x} - e^{-2x}| + C$ , where $C$ is an arbitrary constant.

$\frac{1}{2}\log_e|e^{2x} + e^{-2x}| + C$ , where $C$ is an arbitrary constant.

$\log_e|e^{2x} + e^{-2x}| + C$ , where $C$ is an arbitrary constant.

Correct Answer

Option 3

Correct Answer

Explanation

Q11:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Core

$\int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx$ is equal to

$\frac{1}{2}\log_e 2 - 1$

$2\log_e 2$

$2\log_e 2 - 2$

$4\log_e 2$

Correct Answer

Option 2

Correct Answer

Explanation

Q12:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Core

$\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$ is equal to

$\frac{5\sqrt{2}}{3}$

$\frac{4\sqrt{2}}{3}$

$\frac{4\sqrt{3}}{3}$

$\frac{\sqrt{2}}{3}$

Correct Answer

Option 2

Correct Answer

Explanation

Q13:

2025: 2 June Shift 1

Calculus

>

Integrals

Easy

Core

The value of $\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx$ is

0

-1

$\pi$

1

Correct Answer

Option 1

Correct Answer

Explanation

Q14:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Applied

The value of $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$ is:

$a$

$2a$

$\frac{a}{2}$

$7a$

Correct Answer

Option 1

Correct Answer

Explanation

Q15:

2025: 2 June Shift 1

Calculus

>

Integrals

Medium

Applied

$\int e^{(x \log 5)}e^x dx$ , is:

Where $C$ is the constant of integration.

$\frac{5^x}{\log 5}e^x + C$

$5^x e^x + C$

$(5e)^x \log(5e) + C$

$\frac{(5e)^x}{\log(5e)} + C$

Correct Answer

Option 4

Correct Answer

Explanation

Q16:

2025: 30 May Shift 2

Calculus

>

Integrals

Medium

Common

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

$\frac{1}{5}\log_e \left|\frac{x^5-1}{x^5}\right| + C$ ; C is an arbitrary constant

$\log_e \left|\frac{x^5-1}{x^5}\right| + C$ ; C is an arbitrary constant

$\frac{1}{5}(\log_e|x| + \log_e|x^5-1|) + C$ ; C is an arbitrary constant

$\log_e \left|\frac{x}{x^5-1}\right| + C$ ; C is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

Q17:

2025: 30 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int\limits_{\sqrt{log_e 2}}^{\sqrt{log_e 4}} xe^{x^2} dx$ is equal to

$\frac{1}{2}$

1

2

4

Correct Answer

Option 2

Correct Answer

Explanation

Q18:

2025: 30 May Shift 2

Calculus

>

Integrals

Medium

Core

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin|x| + \cos|x|)dx$ , is equal to:

0

1

2

4

Correct Answer

Option 4

Correct Answer

Explanation

Q19:

2025: 30 May Shift 2

Calculus

>

Integrals

Hard

Core

$\int \frac{\sqrt{16+(\log x)^2}}{x} dx$ is equal to (where C is an arbitrary constant)

$2\log\left|\log x + \sqrt{16+(\log x)^2}\right| + \log x\sqrt{16+(\log x)^2} + C$

$16\log\left|\log x + \sqrt{16+(\log x)^2}\right| + \frac{\log x}{2}\sqrt{16+(\log x)^2} + C$

$8\log\left|\log x + \sqrt{16+(\log x)^2}\right| + \frac{\log x}{2}\sqrt{16+(\log x)^2} + C$

$4\log\left|\log x + \sqrt{16+(\log x)^2}\right| + \frac{\log x}{2}\sqrt{16+(\log x)^2} + C$

Correct Answer

Option 3

Correct Answer

Explanation

Q20:

2025: 30 May Shift 2

Calculus

>

Integrals

Medium

Core

Value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$ is

$\frac{\pi}{4}$

$\frac{\pi}{12}$

0

$\pi$

Correct Answer

Option 3

Correct Answer

Explanation

Q21:

2025: 30 May Shift 2

Calculus

>

Integrals

Easy

Applied

$\int_1^{\sqrt{3}} \frac{1}{1+x^2} dx$ is equal to:

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$\frac{\pi}{6}$

$\frac{\pi}{12}$

Correct Answer

Option 4

Correct Answer

Explanation

Q22:

2025: 30 May Shift 2

Calculus

>

Integrals

Medium

Applied

$\int (x^4 + x^2 + 1)d(x^2)$ is equal to: (where c is an integration constant)

$x^6 + x^4 + x^2 + C$

$\frac{x^6}{6} + \frac{x^4}{4} + \frac{x^2}{2} + C$

$\frac{x^6}{3} + \frac{x^4}{2} + x^2 + C$

$x^6 + x^5 + x^4 + x^3 + x^2 + x + C$

Correct Answer

Option 3

Correct Answer

Explanation

Q23:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int \frac{e^{7\log_e x} - e^{6\log_e x}}{e^{4\log_e x} - e^{3\log_e x}} dx$ is equal to:

(Here, c is an arbitrary constant)

$-\frac{x}{2} + c$

$\frac{x^2}{2} + c$

$\frac{x^3}{3} + c$

$\frac{x^4}{4} + c$

Correct Answer

Option 4

Correct Answer

Explanation

Q24:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Common

$\int_2^5 |x - 3|dx$ equals

$\frac{3}{2}$

$\frac{5}{2}$

$\frac{1}{2}$

5

Correct Answer

Option 2

Correct Answer

Explanation

Q25:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Core

Match List-I with List-II

List-I	List-II
(A) $\int_0^1 \frac{x^2}{1 + x^3} dx$	(I) 0
(B) $\int_0^\pi 3\sin x dx$	(II) $2\log_e\left(\frac{3}{2}\right)$
(C) $\int_{-1}^1 \sin^5 x \cos^6 x dx$	(III) 6
(D) $\int_2^3 \frac{4}{x^2 - 1} dx$	(IV) $\frac{1}{3}\log_e 2$

Choose the correct answer from the options given below:

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

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

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

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

Correct Answer

Option 3

Correct Answer

Explanation

Q26:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx =$

(Given that $c$ is an arbitrary constant)

$2\sin x + 2x\cos\alpha + c$

$2\cos x + 2x\sin\alpha + c$

$x\cos x + 2\sin\alpha + c$

$x\cos x + x\sin\alpha + c$

Correct Answer

Option 1

Correct Answer

Explanation

Q27:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Core

$\int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ is equal to

$\frac{\pi}{2}$

$\frac{\pi}{4}$

0

$-\frac{\pi}{2}$

Correct Answer

Option 2

Correct Answer

Explanation

Q28:

2025: 30 May Shift 1

Calculus

>

Integrals

Medium

Applied

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

$\log_e\left|\frac{x + 2}{x + 1}\right| + C$ : where $C$ is an arbitrary constant

$\log_e\left|\frac{x + 1}{x + 2}\right| + C$ : where $C$ is an arbitrary constant

$\log_e|(x + 1)(x + 2)| + C$ : where $C$ is an arbitrary constant

$\log_e|2x + 3| + C$ : where $C$ is an arbitrary constant

Correct Answer

Option 2

Correct Answer

Explanation

Q29:

2025: 29 May Shift 2

Calculus

>

Integrals

Medium

Common

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

0

1

$\frac{1}{2}$

2

Correct Answer

Option 2

Correct Answer

Explanation

Q30:

2025: 29 May Shift 2

Calculus

>

Integrals

Medium

Common

$\int (e^{x\log a} + e^{a\log x}) dx$ is equal to (where $a > 1$ )

$\frac{a^x}{\log a} + \frac{x^{a+1}}{a+1} + C$ ; C is an arbitrary constant

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

$\frac{a^x}{a+1} + \frac{x^{a+1}}{\log a} + C$ ; C is an arbitrary constant

$(a + 1)a^x + (\log a)x^{a+1} + C$ ; C is an arbitrary constant

Correct Answer

Option 1

Correct Answer

Explanation

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