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

Answer: 4/3((x-1)/(x+2))^(1/4) + C. Solution: Write the denominator as (x+2)^2((x-1)/(x+2))^(3/4). Put t=(x-1)/(x+2), so dt=3/((x+2)^2)dx. Integral =1/3 t^(-3/4)dt=4/3t^(1/4)+C=4/3((x-1)/(x+2))^(1/4)+C.

Calculus

Integrals

Hard

common

$\displaystyle\int \dfrac{1}{(x-1)^{3/4}(x+2)^{5/4}}\, dx$ is equal to :

[where C is an arbitrary constant]

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

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

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

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

✅ Correct Option: 2

Write the denominator as $(x+2)^2\left(\dfrac{x-1}{x+2}\right)^{3/4}$ .

Put $t=\dfrac{x-1}{x+2}$ , so $dt=\dfrac{3}{(x+2)^2}dx$ .

Integral $=\dfrac{1}{3}\int t^{-3/4}dt=\dfrac{4}{3}t^{1/4}+C=\dfrac{4}{3}\left(\dfrac{x-1}{x+2}\right)^{1/4}+C$ .

2025: 16 May Shift 1

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:

2026: 19 May Shift 1

The value of $\int_{0}^{1} \frac{e^x}{1+e^{2x}}\,dx$ is

2025: 3 June Shift 1

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:

2022: 6 Aug Shift 2

The value of $\int_{-4}^{4} \log_e \left( \frac{1-x}{1+x} \right) dx$ is equal to

2026: 30 May Shift 1

$\int e^x\left(\log x+\frac{1}{x^2}\right)dx$ is equal to: (consider $\log x=\log_e x$ )

[C is an arbitrary constant]

2025: 26 May Shift 2

$\int \frac{(x-1)e^x}{x^2} dx, x > 0$ equals (where C is an arbitrary constant)

∫1(x−1)3/4(x+2)5/4 dx\displaystyle\int \dfrac{1}{(x-1)^{3/4}(x+2)^{5/4}}\, dx∫(x−1)3/4(x+2)5/41​dx is equal to : [where C is an arbitrary constant]

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\displaystyle\int \dfrac{1}{(x-1)^{3/4}(x+2)^{5/4}}\, dx$ is equal to :

[where C is an arbitrary constant]