I=( x)^2dx is equal to :(Consider _e x= x) Option 1: x( x)^2+2x x+2x+C, where C is an arbitrary constant Option 2: x x+2x x+2x+C, where C is an arbitrary constant Option 3: x( x)^2-2x x+2x+C, where C is an arbitrary constant Option 4: x( x)^2-2x x-2x+C, where C is an arbitrary constant

The answer is 'x( x)^2-2x x+2x+C, where C is an arbitrary constant'. View full solution:

CUET Mathematics - I=( x)^2dx is equal to :(Consider _e x= x) | PYQs + Solutions

CUET Mathematics 2026 22nd May Shift 1

Calculus

Medium

applied

$x(\log x)^2+2x\log x+2x+C$ , where C is an arbitrary constant

$x\log x+2x\log x+2x+C$ , where C is an arbitrary constant

$x(\log x)^2-2x\log x+2x+C$ , where C is an arbitrary constant

$x(\log x)^2-2x\log x-2x+C$ , where C is an arbitrary constant

✅ Correct Option: 3

Related questions:

29 May Shift 2

Match List-I with List-II

List-I	List-II
Integral	Solution: C is an arbitrary constant
---	---
(A) $\int \frac{dx}{x^2 + 25}$	(I) $\frac{1}{10} \log \left\vert \frac{5 + x}{5 - x} \right\vert + C$
(B) $\int \frac{dx}{x^2 - 25}$	(II) $\log \vert x + \sqrt{x^2 - 25}\vert + C$
(C) $\int \frac{dx}{25 - x^2}$	(III) $\frac{1}{5} \tan^{-1} \left( \frac{x}{5} \right) + C$
(D) $\int \frac{dx}{\sqrt{x^2 - 25}}$	(IV) $\frac{1}{10} \log \left\vert \frac{x - 5}{x + 5} \right\vert + C$

Choose the correct answer from the options given below:

21st May Shift 1

The value of $\int_{-2}^{3} |x-1|\, dx$ is

30 May Shift 1

$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx =$

(Given that $c$ is an arbitrary constant)