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.

$$I(a)=\int_0^\infty\frac{x^{-a}}{x^2+2x+2}dx\overset{x\to 1/x}{=}\int_0^\infty\frac{x^a}{2x^2+2x+1}dx$$
$$=\Im\int_0^\infty\frac{(1+i)x^a}{1+(1+i)x}dx\overset{(1+i)x=u}{=}\Im \frac{1}{(1+i)^a}\int_0^\infty\frac{u^a}{1+u}du$$
$$=-\Im\frac{\pi\csc(a\pi)}{(1+i)^a}=2^{-a/2}\csc(a\pi)\sin(\frac{\pi}{4}a),$$
and your integral is $I''(0).$
