Tricky integral

$I = \displaystyle \int_{0}^{\infty} \dfrac{\ln(1+x^2)}{1+x^2} dx$

We make a substitution,

$x = \tan t \Rightarrow dx = \sec^2t dt$

Hence the limits change to $0$ and $\dfrac{\pi}{2}$

$I = \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\sec^2t)}{\sec^2t} \sec^2t dt$

$I = \displaystyle \int_{0}^{\frac{\pi}{2}} (-2)\ln \cos t dt$

$\displaystyle \int_{0}^{\frac{\pi}{2}} \ln \cos t dt = \dfrac{\pi}{-2} \ln 2$

Hence,

$I = \displaystyle (-2) \dfrac{\pi}{-2} \ln 2 \Rightarrow I = \pi \ln 2$

just sub x=tanu..then u get a very popular integral