JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

The value of the integral $\int_{0}^{\infty} \frac{lo g _{e} ( x )}{x ^{2} + 4} d x$ is :

A
$\frac{π lo g _{e} 2}{2}$
B
$\frac{π lo g _{e} 2}{4}$
C
$1 + π lo g_{e} 2$
D
$2 + π lo g_{e} 2$

Solution

Substitute $x = 2 t$ , $d x = 2 d t$ :

$I = \int_{0}^{\infty} \frac{ln 2 + ln t}{4 t ^{2} + 4} \cdot 2 d t = \frac{ln 2}{2} \int_{0}^{\infty} \frac{d t}{t ^{2} + 1} + \frac{1}{2} \int_{0}^{\infty} \frac{ln t}{t ^{2} + 1} d t$ .

The first integral is $\frac{π}{2}$ . For the second, substitute $t = 1/ u$ : $\int_{0}^{\infty} \frac{ln t}{t ^{2} + 1} d t = - \int_{0}^{\infty} \frac{ln u}{u ^{2} + 1} d u$ , so the integral equals its negative and must be $0$ .

Hence $I = \frac{ln 2}{2} \cdot \frac{π}{2} = \frac{π ln 2}{4}$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →