The value of the integral $\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:

Let $I = \displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ Substitute $x = 2 \tan \theta$, which gives $dx = 2 \sec^2 \theta \,d\theta$. The limits of integration change from $x = 0 \to \theta = 0$ to $x \to \infty \to \theta = \dfrac{\pi}{2}$. $I = \displaystyle\int_0^{\pi/2} \dfrac{\log_e(2 \tan \theta)}{4 \tan^2 \theta + 4} (2 \sec^2 \theta) \,d\theta$ $I = \displaystyle\int_0^{\pi/2} \dfrac{\log_e(2 \tan \theta)}{4 \sec^2 \theta} (2 \sec^2 \theta) \,d\theta$ $I = \dfrac{1}{2} \displaystyle\int_0^{\pi/2} \log_e(2 \tan \theta) \,d\theta$ $I = \dfrac{1}{2} \displaystyle\int_0^{\pi/2} (\log_e 2 + \log_e(\tan \theta)) \,d\theta$ $I = \dfrac{1}{2} \log_e 2 \displaystyle\int_0^{\pi/2} 1 \,d\theta + \dfrac{1}{2} \displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta$ Let $I_1 = \displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta$ Using the property $\displaystyle\int_0^a f(x)\,dx = \displaystyle\int_0^a f(a-x)\,dx$, we get: $I_1 = \displaystyle\int_0^{\pi/2} \log_e\left(\tan\left(\dfrac{\pi}{2} - \theta\right)\right) \,d\theta = \displaystyle\int_0^{\pi/2} \log_e(\cot \theta) \,d\theta$ $I_1 = \displaystyle\int_0^{\pi/2} \log_e\left(\dfrac{1}{\tan \theta}\right) \,d\theta = -\displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta = -I_1$ $2I_1 = 0 \Rightarrow I_1 = 0$ Substituting $I_1 = 0$ back into the equation for $I$: $I = \dfrac{1}{2} \log_e 2 \left[\theta\right]_0^{\pi/2} + 0$ $I = \dfrac{1}{2} \log_e 2 \left(\dfrac{\pi}{2}\right) = \dfrac{\pi \log_e 2}{4}$ Answer: $\dfrac{\pi\log_e(2)}{4}$

JEE Main 2026 — Mathematics Calculus

2026mcqmedium

The value of the integral $\displaystyle \displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\dfrac{\pi\log_e(2)}{2}$
B
$\dfrac{\pi\log_e(2)}{4}$
C
$1 + \pi\log_e(2)$
D
$2 + \pi\log_e(2)$

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Solution

Let $\displaystyle I = \displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$

Substitute $x = 2 \tan \theta$ , which gives $dx = 2 \sec^2 \theta \,d\theta$ .

The limits of integration change from $x = 0 \to \theta = 0$ to $x \to \infty \to \theta = \dfrac{\pi}{2}$ .

$I = \displaystyle\int_0^{\pi/2} \dfrac{\log_e(2 \tan \theta)}{4 \tan^2 \theta + 4} (2 \sec^2 \theta) \,d\theta$

$I = \displaystyle\int_0^{\pi/2} \dfrac{\log_e(2 \tan \theta)}{4 \sec^2 \theta} (2 \sec^2 \theta) \,d\theta$

$I = \dfrac{1}{2} \displaystyle\int_0^{\pi/2} \log_e(2 \tan \theta) \,d\theta$

$I = \dfrac{1}{2} \displaystyle\int_0^{\pi/2} (\log_e 2 + \log_e(\tan \theta)) \,d\theta$

$I = \dfrac{1}{2} \log_e 2 \displaystyle\int_0^{\pi/2} 1 \,d\theta + \dfrac{1}{2} \displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta$

Let $\displaystyle I_1 = \displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta$

Using the property $\displaystyle \displaystyle\int_0^a f(x)\,dx = \displaystyle\int_0^a f(a-x)\,dx$ , we get:

$I_1 = \displaystyle\int_0^{\pi/2} \log_e\left(\tan\left(\dfrac{\pi}{2} - \theta\right)\right) \,d\theta = \displaystyle\int_0^{\pi/2} \log_e(\cot \theta) \,d\theta$

$I_1 = \displaystyle\int_0^{\pi/2} \log_e\left(\dfrac{1}{\tan \theta}\right) \,d\theta = -\displaystyle\int_0^{\pi/2} \log_e(\tan \theta) \,d\theta = -I_1$

$2I_1 = 0 \Rightarrow I_1 = 0$

Substituting $I_1 = 0$ back into the equation for $I$ :

$I = \dfrac{1}{2} \log_e 2 \left[\theta\right]_0^{\pi/2} + 0$

$I = \dfrac{1}{2} \log_e 2 \left(\dfrac{\pi}{2}\right) = \dfrac{\pi \log_e 2}{4}$

Answer: $\dfrac{\pi\log_e(2)}{4}$

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