Letting $x=\tan \theta$ yields $$ I:=\int_{0}^{\infty} \frac{\ln \left(1+x^{2}\right)}{1+x^{2}} d x= \int_{0}^{\frac{\pi}{2}} \ln \left(\sec ^{2} \theta\right) d \theta =-2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta $$ By my post, $$ \begin{aligned} I &=-2\left(-\frac{\pi}{2} \ln 2\right)=\pi \ln 2 \end{aligned} $$
My question: Is there any other simpler method to find the integral?
