Calculus

Integrals

Easy

core

The value of $\int_{0}^{1} \frac{e^x}{1+e^{2x}}\,dx$ is

$\tan e - \frac{\pi}{4}$

$\tan^{-1}e - \frac{\pi}{4}$

$\tan^{-1}e - \frac{\pi}{2}$

$\tan^{-1}e - \frac{\pi}{3}$

✅ Correct Option: 2

Substitute $t = e^{x}$ , so $dt = e^{x}dx$ and limits become 1 to $e$ . The integral is $\int_1^e \frac{dt}{1+t^2} = \tan^{-1}e - \tan^{-1}1 = \tan^{-1}e - \frac{\pi}{4}$ .

2025: 13 May Shift 1

$\int \frac{x^3 - 1}{x^2} dx$ is equal to

2026: 26 May Shift 2

$\displaystyle\int \sqrt[3]{x}(x^3+4x^2+7)\,dx$ is equal to-

2026: 29 May Shift 2

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

2025: 3 June Shift 1

The value of $\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to (where C is an arbitrary constant)

2022: 30 Aug Shift 1

$\int \sqrt{1 - 49x^2} \, dx$ is equal to

2026: 18 May Shift 1

If $\int \frac{1}{\tan x + \cot x}dx = k\cos 2x + C$ , where C is an arbitrary constant, then the value of $k$ is

The value of ∫01ex1+e2x dx\int_{0}^{1} \frac{e^x}{1+e^{2x}}\,dx∫01​1+e2xex​dx is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The value of $\int_{0}^{1} \frac{e^x}{1+e^{2x}}\,dx$ is