xsqrt(x^2-1) dx is equal to : ( where C is an arbitrary constant ) Option 1: 1sqrt(x^2-1)+C Option 2: 2/3(x^2-1)^3/2+C Option 3: 1/3(x^2-1)^3/2+C Option 4: -1sqrt(x^2-1)+C

The answer is '1/3(x^2-1)^3/2+C'. View full solution:

CUET Mathematics - xsqrt(x^2-1) dx is equal to : ( where C is an arbitrary constant ) | PYQs + Solutions

CUET Mathematics 2026 26th May Shift 2

Calculus

Easy

common

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

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

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

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

✅ Correct Option: 3

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Match List-I with List-II

List-I	List-II
Functions	Integrals
---	---
(A) $\int \frac{dx}{x^2 - 4}, x \neq \pm 2$	(I) $\log \vert x + \sqrt{4 + x^2}\vert + C$ , where $C$ is an arbitrary constant
(B) $\int \frac{1}{\sqrt{16 - x^2}} dx; \vert x\vert < 4$	(II) $\sin^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant
(C) $\int \frac{1}{16 + x^2} dx$	(III) $\frac{1}{4} \log \left\vert \frac{x-2}{x+2} \right\vert + C$ , where $C$ is an arbitrary constant
(D) $\int \frac{1}{\sqrt{4 + x^2}} dx$	(IV) $\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$ , where $C$ is an arbitrary constant

Choose the correct answer from the options given below:

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