$I=\int_0^{\infty}\!\frac{\ln\left(\tan\left(\tan^{-1}\left(e^{\frac{1}{\pi}\arctan u}\right)\right)\right)}{u^2+2\pi u+2\pi^2}\,du$
$=\int_0^{\infty}\!\frac{\ln\left(e^{\frac{1}{\pi}\arctan u}\right)}{u^2+2\pi u+2\pi^2}\,du$
$=\int_0^{\infty}\!\frac{\left(\frac{1}{\pi}\arctan u\right)\ln\left(e\right)}{u^2+2\pi u+2\pi^2}\,du$
$=\frac{1}{\pi}\int_0^{\infty}\!\frac{\arctan u}{u^2+2\pi u+2\pi^2}\,du$
$=\frac{1}{\pi}\int_0^{\infty}\!\frac{\arctan u}{\left(u+\pi\right)^2+\pi^2}\,du$
[ Note: $tan^{-1}u=\arctan u$
The integral part is a standard type integral with known result:
$\int_0^{\infty}\!\frac{\arctan u}{\left(u+a\right)^2+b^2}\,du$
$=\frac{\pi}{2}\int_0^{\infty}\!\frac{\arctan\left(\frac{a}{b}\right)}{b}\,du$
[ $f$ or $b>0$ ]
[ In our case : $a=-\pi$, $b=\pi$ ]
$\Rightarrow\frac{1}{\pi}\int_0^{\infty}\!\frac{\arctan u}{\left(u+\pi\right)^2+\pi^2}\,du$
$=\frac{1}{\pi}\cdot\frac{\pi}{2}\cdot\frac{\arctan\left(-1\right)}{\pi}$
$=\frac{1}{\pi}\cdot\frac{\pi}{2}\cdot\frac{\left(-\frac{\pi}{4}\right)}{\pi}$
$=-\frac18$
[Credit: Patra N]