The value of the integral I=^x(x-2)/(x+1)^4dx is, [C is an arbitrary constant] Option 1: e^x/(x+1)^4+C Option 2: e^x/(x+1)^3+C Option 3: -3e^x/(x+1)^4+C Option 4: 3e^x/(x+1)^3+C

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

CUET Mathematics - The value of the integral I=^x(x-2)/(x+1)^4dx is, [C is an arbitrary constant] | PYQs + Solutions

CUET Mathematics 2026 12th May Shift 2

Calculus

Medium

applied

$\dfrac{e^x}{(x+1)^4}+C$

$\dfrac{e^x}{(x+1)^3}+C$

$\dfrac{-3e^x}{(x+1)^4}+C$

$\dfrac{3e^x}{(x+1)^3}+C$

✅ Correct Option: 2

Related questions:

12th May Shift 2

Match List-I with List-II

List-I	List-II
Indefinite integral	Solution (where 'c' is an arbitrary constant)
(A) $\displaystyle\int \sqrt{x^2-a^2} \, dx$	(I) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + c$
(B) $\displaystyle\int \dfrac{1}{x^2-a^2} \, dx$	(II) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left[x+\sqrt{x^2+a^2}\right] + c$
(C) $\displaystyle\int \dfrac{1}{x^2+a^2} \, dx$	(III) $\dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left[x+\sqrt{x^2-a^2}\right] + c$
(D) $\displaystyle\int \sqrt{x^2+a^2} \, dx$	(IV) $\dfrac{1}{2a}\log\left\vert \dfrac{x-a}{x+a} \right\vert + C$

Choose the correct answer from the options given below:

21st May Shift 2

The value of $I = \displaystyle\int (x^3-1)^{1/3}x^5\,dx$ is

29th May Shift 2

$\int_1^2 \frac{dx}{(x+1)(x+2)} =$