_(-1)^1(x+√(x^2+1))dx is equal to Option 1: -2 Option 2: 2 Option 3: 0 Option 4: 1/22

Answer: 0. Solution: Let f(x)=(x+√(x^2+1)). Then f(-x)=1/(x+√(x^2+1))=-f(x), so f is odd. The integral of an odd function over the symmetric interval [-1,1] is 0.

Calculus

Integrals

Easy

core

$\int_{-1}^{1}\log(x+\sqrt{x^2+1})dx$ is equal to

$-\log2$

$\log2$

$\frac{1}{2}\log2$

✅ Correct Option: 3

Let $f(x)=\log\left(x+\sqrt{x^2+1}\right)$ . Then $f(-x)=\log\frac{1}{x+\sqrt{x^2+1}}=-f(x)$ , so $f$ is odd.

The integral of an odd function over the symmetric interval $[-1,1]$ is $0$ .

2025: 13 May Shift 2

$\int \left(\frac{1}{log_e t} - \frac{1}{(log_e t)^2}\right) dt$ is equal to

2025: 3 June Shift 2

$\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$ is equal to

2025: 3 June Shift 2

Match List-I with List-II

List-I	List-II
(A) $\int \dfrac{dx}{x^2 - 16}$	(I) $\dfrac{1}{8} \log \left\vert \dfrac{4 + x}{4 - x} \right\vert + c$ , Where C is an arbitrary constant,
(B) $\int \dfrac{dx}{x^2 + 16}$	(II) $\log \left\vert x + \sqrt{x^2 - 16} \right\vert + c$ , Where C is an arbitrary constant,
(C) $\int \dfrac{dx}{16 - x^2}$	(III) $\dfrac{1}{8} \log \left\vert \dfrac{x - 4}{x + 4} \right\vert + c$ , Where C is an arbitrary constant,
(D) $\int \dfrac{dx}{\sqrt{x^2 - 16}}$	(IV) $\dfrac{1}{4} \tan^{-1} \left( \dfrac{x}{4} \right) + c$ , Where C is an arbitrary constant,

Choose the correct answer from the options given below:

2026: 26 May Shift 2

Value of $\displaystyle\int_{-1}^{1} \dfrac{x^5+|x|+1}{x^2+2|x|+1}\,dx$ is equal to :

2026: 26 May Shift 2

$\displaystyle\int \dfrac{1-\sin x}{\cos^2 x}\,dx$ is equal to:

(where $c$ is an arbitrary constant)

2026: 31 May Shift 2

The value of $\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx$ is:

∫−11log⁡(x+x2+1)dx\int_{-1}^{1}\log(x+\sqrt{x^2+1})dx∫−11​log(x+x2+1​)dx is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int_{-1}^{1}\log(x+\sqrt{x^2+1})dx$ is equal to