<s>[INST] Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx 
 + \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\ 
 &= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
 -2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}
which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.
It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.
 [/INST] Here is a solution based on real methods. Note that $\frac{1}{(x+1)^2+1}=\frac{x^2-2x+2}{x^4+4}$. Then
\begin{align}
&\int_0^\infty \frac{\ln^2x}{(x+1)^2+1}dx \\
= & \int_0^\infty {\frac{x^2\ln^2x }{x^4+4}}\overset{x^2=1/t^2}{dx}
-2\int_0^\infty {\frac{x\ln^2x }{x^4+4}} \overset{x^2=2t}{dx} 
 + 2\int_0^\infty { \frac{\ln^2x }{x^4+4} } \overset{x^2=2t^2}{dx} \\
= &
-\frac18\ln^22\int_0^\infty \frac{1}{t^2+1}dt
 -\frac18\int_0^\infty \frac{\ln^2t}{t^2+1}dt \\
& +\frac{1}{2\sqrt2} \ln^22  \int_0^\infty \frac{1}{t^4+1}dt
 + \sqrt2 \int_0^\infty \frac{\ln^2 t}{t^4+1}dt \\
= & 
-\frac{\ln^22}8\left(\frac\pi2\right)
 -\frac18\left(\frac{\pi^3}8\right) +\frac{\ln^22 }{2\sqrt2}\left(\frac\pi{2\sqrt2}\right)
 + \sqrt2 \left(\frac{3\pi^3\sqrt2}{64}\right) \\
=& \frac{5\pi^3}{64}+\frac\pi{16}\ln^22
\end{align}
</s>