The value of _(-4)^4 _e ( (1-x)/(1+x) ) dx is equal to Option 1: 2 Option 2: 1 Option 3: 0 Option 4: 8 _e(2)

Answer: 0. Solution: Let f(x) = _e (1-x)/(1+x). f(-x) = _e (1+x)/(1-x) = -f(x), so f is odd. Integral of an odd function over a symmetric interval is zero.

Calculus

Integrals

Easy

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

$2$

$1$

$0$

$8 \log_e(2)$

✅ Correct Option: 3

Let $f(x) = \log_e \frac{1-x}{1+x}$ .

$f(-x) = \log_e \frac{1+x}{1-x} = -f(x)$ , so $f$ is odd.

Integral of an odd function over a symmetric interval is zero.

2022: 30 Aug Shift 1

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

2026: 22 May Shift 2

$\int \left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\} dx$ is equal to:(where C is an arbitrary constant)

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:

2025: 30 May Shift 2

$\int \frac{\sqrt{16+(\log x)^2}}{x} dx$ is equal to (where C is an arbitrary constant)

2022: 10 Aug Shift 1

$\int_0^{4\pi} \frac{x}{1 + |\cos x|} dx =$

2025: 3 June Shift 1

$\int \sqrt{1 + \frac{x^2}{9}} dx$ is equal to

(Where C is an arbitrary constant)

The value of ∫−44log⁡e(1−x1+x)dx\int_{-4}^{4} \log_e \left( \frac{1-x}{1+x} \right) dx∫−44​loge​(1+x1−x​)dx is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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