<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] We can use the integral of
\begin{equation}
  f(z)=\frac{\ln^3z}{(z+1)^2+2}
 \end{equation}
along the classical keehole contour: above the positive real axis , along the large circle, back from below along the positive axis and avoiding the origin by a small circle. Except the horizontal parts, the other contributions can be shown to vanish. Above the axis, $z=x+i0^+$ and $\ln^3(z)=\ln^3(x)$, while below, $z=x+i0^-$ and $\ln^3(z)=(\ln(x)+2i\pi)^3$. The integral can then be written as
\begin{align}
  J&=\int_0^\infty\frac{\ln^3x}{(x+1)^2+1}dx-\int_0^\infty\frac{(\ln(x)+2i\pi)^3}{(x+1)^2+1}dx
 \end{align}
Considering the imaginary part of the integral,
\begin{align}
 \Im J&=-3 \left( 2\pi \right)\int_0^\infty\frac{\ln^2x}{(x+1)^2+1}dx+\left( 2\pi \right)^3\int_0^\infty\frac{1}{(x+1)^2+1}dx\\
 \Im J&=-6\pi I+2\pi^4
\end{align}
The poles of the function are $z_{\pm}=-1\pm i$, or $z_+=\sqrt{2}e^{i3\pi/4},z_-=\sqrt{2}e^{i5\pi/4}$. Corresponding residues $R_\pm=\ln^3(z_\pm)/\left( 2(1+z_\pm) \right)$, we deduce
\begin{equation}
 J=2i\pi\left[\frac{\left( \ln\sqrt{2} +3i\pi/4\right)^3}{2i}-\frac{\left( \ln\sqrt{2} +5i\pi/4\right)^3}{2i}\right]
\end{equation}
Then,
$$\Im J=\frac{49\pi^4}{32}-\frac{3\pi^2}{8}$$
Finally,
\begin{equation}
 I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx = \frac{5\pi^3}{64}+\frac\pi{16}\ln^22
\end{equation}
</s>