The value of $\int\limits_0^{\infty} \frac{\log x}{1+x^2} d x$, is

0

Let $I=\int\limits_0^{\infty} \frac{\log x}{1+x^2} d x$

Putting $x=\tan \theta$, we get

$I=\int\limits_0^{\pi / 2} \log \tan \theta d \theta$   ...(1)

$I=\int\limits_0^{\pi / 2} \log \tan(\frac{\pi}{2}-θ)dθ=\int\limits_0^{\pi / 2}\log \cot θdθ$   ...(2)

Eq. (1) + Eq. (2)

$2I=\int\limits_0^{\pi / 2}\log \tan θ\cot θdθ=\int\limits_0^{\pi / 2}0dθ=0$

$I=0$