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

The answer is '4/3(x-1/x+2)^1/4 + C'. View full solution:

CUET Mathematics - 1(x-1)^3/4(x+2)^5/4 dx is equal to : [where C is an arbitrary constant] | PYQs + Solutions

CUET Mathematics 2026 25th May Shift 1

Calculus

Hard

common

$\dfrac{3}{4}\left(\dfrac{x-1}{x+2}\right)^{3/4} + C$

$\dfrac{4}{3}\left(\dfrac{x-1}{x+2}\right)^{1/4} + C$

$\dfrac{4}{3}\left(\dfrac{x-1}{x+2}\right)^{-1/4} + C$

$\dfrac{3}{4}\left(\dfrac{x+2}{x-1}\right)^{3/4} + C$

✅ Correct Option: 2

Related questions:

21 May Shift 1

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

List-I	List-II
(A) $\int \frac{dx}{\sqrt{a^2 - x^2}} =$	(I) $\log_e \vert x + \sqrt{x^2 - a^2}\vert + c$
(B) $\int \sqrt{a^2 - x^2} dx =$	(II) $\sin^{-1} \frac{x}{a} + c$
(C) $\int \sqrt{x^2 - a^2} dx =$	(III) $\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + c$
(D) $\int \frac{dx}{\sqrt{x^2 - a^2}} =$	(IV) $\frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \log_e \vert x + \sqrt{x^2 - a^2}\vert + c$

Choose the correct answer from the options given below:

19 May Shift 1

Match List-I with List-II

Choose the correct answer from the options given below:

22nd May Shift 1

$\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx$ is equal to : (Consider $\log_e x = \log x$ )