x^8/((x^3+1)^(1/3)) dx is equal to Option 1: 1/4(x^3+1)^(4/3) - 1/2(x^3+1)^(2/3) - 2/5(x^3+1)^(1/3) + C, C is an arbitrary constant Option 2: 1/4(x^3+1)^(4/3) + 1/2(x^3+1)^(2/3) - 2/5(x^3+1)^(1/3) + C, C is an arbitrary constant Option 3: 1/8(x^3+1)^(8/3) - 1/2(x^3+1)^(2/3) + 2/5(x^3+1)^(5/3) + C, C is an arbitrary constant Option 4: 1/8(x^3+1)^(8/3) + 1/2(x^3+1)^(2/3) - 2/5(x^3+1)^(5/3) + C, C is an arbitrary constant

Answer: 1/8(x^3+1)^(8/3) + 1/2(x^3+1)^(2/3) - 2/5(x^3+1)^(5/3) + C, C is an arbitrary constant. Solution: Put u=x^3+1, so du=3x^2 dx and x^6=(u-1)^2. The integral becomes 1/3(u^(5/3)-2u^(2/3)+u^(-1/3))du. Integrating term by term gives 1/8u^(8/3)-2/5u^(5/3)+1/2u^(2/3)+C, i.e. option 4 with u=x^3+1.

Calculus

Integrals

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$\int \frac{x^8}{(x^3+1)^{1/3}} dx$ is equal to

$\frac{1}{4}(x^3+1)^{4/3} - \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{4}(x^3+1)^{4/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{1/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} - \frac{1}{2}(x^3+1)^{2/3} + \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

$\frac{1}{8}(x^3+1)^{8/3} + \frac{1}{2}(x^3+1)^{2/3} - \frac{2}{5}(x^3+1)^{5/3} + C$ , C is an arbitrary constant

✅ Correct Option: 4

Put $u=x^3+1$ , so $du=3x^2\,dx$ and $x^6=(u-1)^2$ .

The integral becomes $\frac{1}{3}\int\left(u^{5/3}-2u^{2/3}+u^{-1/3}\right)du$ .

Integrating term by term gives $\frac{1}{8}u^{8/3}-\frac{2}{5}u^{5/3}+\frac{1}{2}u^{2/3}+C$ , i.e. option 4 with $u=x^3+1$ .

2022: 16 July Shift 2

The value of $\int_{\pi/2}^{\pi} \frac{1}{1 + \cot x} dx$ is equal to:

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
Definite integral	Value
---	---
(A) $\int_{1}^{e} \frac{\log x}{x} dx$	(I) 4
(B) $\int_{-2}^{2} x^3(1 - x^2) dx$	(II) $\frac{1}{2}$
(C) $\int_{1}^{2} x \, dx$	(III) 0
(D) $\int_{-2}^{2} \vert x\vert dx$	(IV) $\frac{3}{2}$

Choose the correct answer from the options given below:

2025: 21 May Shift 1

The value of $\int \left\{ \frac{1}{\log_e x} - \frac{1}{(\log_e x)^2} \right\} dx$ is

2025: 2 June Shift 1

The value of $\int_{-\pi/2}^{\pi/2}(x^5 + x^3\cos x)dx$ is

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $\displaystyle\int_{\pi/6}^{\pi/3} \cos x \, dx$	(I) $0$
(B) $\displaystyle\int_{-\pi}^{\pi} (\sin^{61}x + x^{123}) \, dx$	(II) $4$
(C) $\displaystyle\int_{0}^{\pi/2} \dfrac{1}{1+\tan^3 x} \, dx$	(III) $\dfrac{\sqrt{3}-1}{2}$
(D) $\displaystyle\int_{-2}^{2} (2 - \left\vert x \right\vert) \, dx$	(IV) $\dfrac{\pi}{4}$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

The value of $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$ is:

∫x8(x3+1)1/3dx\int \frac{x^8}{(x^3+1)^{1/3}} dx∫(x3+1)1/3x8​dx is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \frac{x^8}{(x^3+1)^{1/3}} dx$ is equal to