The value of _0^1 e^x (x + 1) dx is equal to Option 1: e + 1 Option 2: 3e Option 3: e Option 4: 3e - 2

Answer: e. Solution: Note that d/dx[x e^x] = e^x + x e^x = e^x(x+1). So _0^1 e^x(x+1) dx = [x e^x]_0^1 = e - 0 = e.

Calculus

Integrals

Easy

The value of $\int_0^1 e^x (x + 1) \, dx$ is equal to

$e + 1$

$3e$

$e$

$3e - 2$

✅ Correct Option: 3

Note that $\frac{d}{dx}[x e^x] = e^x + x e^x = e^x(x+1)$ .

So $\int_0^1 e^x(x+1) dx = [x e^x]_0^1 = e - 0 = e$ .

2026: 21 May Shift 1

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If $\int (1 + e^{-x} + e^{-2x} + ...)dx = \log\phi(x) + C$ , then $\phi(x)$ is equal to :

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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:

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$\int_{0}^{\frac{\pi}{2}} \frac{1-\cot x}{\operatorname{cosec} x+\cos x} d x=$