_1^e x_ex dx is equal to: Option 1: 1/4(2e^2+1) Option 2: 1/4(e^2+1) Option 3: 1/4(e^2-1) Option 4: 1/4(2e^2-1)

Answer: 1/4(e^2+1). Solution: Integrate by parts with u = _e x, dv = x dx: x_e x dx = x^2/2_e x - x^2/4. Evaluating from 1 to e: (e^2/2-e^2/4) - (0-1/4) = (e^2+1)/4.

Calculus

Integrals

Easy

common

$\displaystyle\int_1^e x\log_ex\,dx$ is equal to:

$\dfrac{1}{4}(2e^2+1)$

$\dfrac{1}{4}(e^2+1)$

$\dfrac{1}{4}(e^2-1)$

$\dfrac{1}{4}(2e^2-1)$

✅ Correct Option: 2

Integrate by parts with $u = \log_e x$ , $dv = x\,dx$ :

$\int x\log_e x\,dx = \dfrac{x^2}{2}\log_e x - \dfrac{x^2}{4}$ .

Evaluating from $1$ to $e$ : $\left(\dfrac{e^2}{2}-\dfrac{e^2}{4}\right) - \left(0-\dfrac{1}{4}\right) = \dfrac{e^2+1}{4}$ .

2026: 15 May Shift 1

Value of $\displaystyle\int \dfrac{dx}{\sin^{\frac{3}{5}}x \cdot \cos^{\frac{7}{5}}x}$ is (where c is an arbitrary constant)

2026: 15 May Shift 1

$\displaystyle\int \dfrac{dx}{(x^2+4)^{\frac{3}{2}}}$ is equal to ( where $c$ is an arbitrary constant)

2026: 22 May Shift 2

Let $f(x)=\int \frac{3x-2}{(x+1)^2(x+3)}dx = \alpha\log\left|\frac{x+1}{x+3}\right|+\beta\left(\frac{1}{x+1}\right)+\gamma$ , where $\gamma$ is the constant of integration. If $f(0)=\frac{11}{4}\log\left(\frac{1}{3}\right)+4$ , then which of the following are correct?

(A) $4\alpha+2\beta+6\gamma=22$

(B) $f(1)=\alpha[1-\log 2]$

(D) $\gamma=-\frac{5}{2}$

Choose the correct answer from the options given below:

2022: 10 Aug Shift 1

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

2025: 14 May Shift 1

Match List-I with List-II

List-I	List-II
Definite integral	Value
(A) $\int_0^1 \frac{2x}{1 + x^2} dx$	(I) 2
(B) $\int_{-1}^1 sin^3 x \cos^4 x dx$	(II) $log_e\left(\frac{3}{2}\right)$
(C) $\int_0^{\pi} \sin x dx$	(III) $log_e 2$
(D) $\int_2^3 \frac{2}{x^2 - 1} dx$	(IV) 0

Choose the correct answer from the options given below:

2026: 30 May Shift 1

The value of $\int_{-\pi/2}^{\pi/2}(\sin|x|+\cos|x|)dx$ equals:

∫1exlog⁡ex dx\displaystyle\int_1^e x\log_ex\,dx∫1e​xloge​xdx is equal to:

Solution

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$\displaystyle\int_1^e x\log_ex\,dx$ is equal to: