Definite IntegrationmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Integral of ln x /(x²+4) from 0 to ∞ = π ln2 / 4 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The value of the integral ∫0∞log⁡e(x)x2+4 dx\displaystyle\int_0^{\infty}\dfrac{\log_e(x)}{x^2+4}\,dx is
Aπlog⁡e(2)2\dfrac{\pi\log_e(2)}{2}
Bπlog⁡e(2)4\dfrac{\pi\log_e(2)}{4}correct
C1+πlog⁡e(2)1+\pi\log_e(2)
D2+πlog⁡e(2)2+\pi\log_e(2)
Solution
Step 1: Let x=2tx=2t, dx=2 dtdx=2\,dt, x2+4=4(t2+1)x^2+4=4(t^2+1), ln⁡x=ln⁡2+ln⁡t\ln x=\ln2+\ln t:
I=∫0∞ln⁡(2t)4(t2+1) (2 dt)=12∫0∞ln⁡2t2+1 dt+12∫0∞ln⁡tt2+1 dt.I=\int_0^{\infty}\frac{\ln(2t)}{4(t^2+1)}\,(2\,dt)=\frac12\int_0^{\infty}\frac{\ln2}{t^2+1}\,dt+\frac12\int_0^{\infty}\frac{\ln t}{t^2+1}\,dt.
 Step 2: ∫0∞dtt2+1=[arctan⁡t]0∞=π2\displaystyle\int_0^{\infty}\frac{dt}{t^2+1}=\big[\arctan t\big]_0^{\infty}=\frac{\pi}{2}. Let J=∫0∞ln⁡tt2+1 dtJ=\displaystyle\int_0^{\infty}\dfrac{\ln t}{t^2+1}\,dt; sub t→1ut\to\dfrac1u:
J=∫0∞−ln⁡uu2+1 du=−J⇒J=0.J=\int_0^{\infty}\frac{-\ln u}{u^2+1}\,du=-J\Rightarrow J=0.
 Step 3: I=12(ln⁡2⋅π2)+0=πln⁡24I=\dfrac12\left(\ln2\cdot\dfrac{\pi}{2}\right)+0=\dfrac{\pi\ln2}{4}. Correct answer: (2)
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