JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q18

MathematicsSingle correct+4 / -1Question 18 of 75

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

\dfrac{\pi\log_e(2)}2

\dfrac{\pi\log_e(2)}4

1+\pi\log_e(2)

2+\pi\log_e(2)

Answer and worked solution

Official answer: B

Substitute $x=4/t$ . $dx=-4\,dt/t^2$ , and
$I=\int_0^\infty\frac{\ln4-\ln t}{t^2+4}\,dt=\ln4\int_0^\infty\frac{dt}{t^2+4}-I.$

So $2I=\ln4\cdot\dfrac\pi4$ , giving $I=\dfrac{\pi\ln2}4$ .

(B)

Topics:Properties of definite integrals

JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 18