sqrt(1+x^2)x^4 dx is equal to : Option 1: (1+1/x^2)^3/2+C :Where C is an arbitrary constant Option 2: -1/3(1+1/x^2)^3/2+C :Where C is an arbitrary constant Option 3: 1/3(1+1/x^2)^3/2+C :Where C is an arbitrary constant Option 4: -1/3(1+1/x^2)^5/2+C :Where C is an arbitrary constant

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

CUET Mathematics - sqrt(1+x^2)x^4 dx is equal to : | PYQs + Solutions

CUET Mathematics 2026 14th May Shift 2

Calculus

Medium

core

$\left(1+\frac{1}{x^2}\right)^{3/2}+C$ :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 Option: 2

	LIST-I Definite Integrals		LIST-II Values
A.	$\displaystyle\int_{-1}^{1} \vert x \vert \, dx$	I.	$2$
B.	$\displaystyle\int_{-1}^{1} \sin^3 x \cos^4 x \, dx$	II.	$1$
C.	$\displaystyle\int_{-1}^{1} 3\vert x \vert^2 \, dx$	III.	$0$
D.	$\displaystyle\int_{0}^{4} \vert x-1 \vert \, dx$	IV.	$5$