JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 76

Put $2x=\tan\theta$, so $dx=\frac12\sec^2\theta\,d\theta$ and $1+4x^2=\sec^2\theta$:
$$I=\frac12\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta=\frac12\cdot\frac\pi8\ln2=\frac\pi{16}\ln2,$$
using the standard result $\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta=\frac\pi8\ln2$.