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.
 A: 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}
A: $$I(a)=\int_0^\infty\frac{x^{-a}}{x^2+2x+2}dx\overset{x\to 1/x}{=}\int_0^\infty\frac{x^a}{2x^2+2x+1}dx$$
$$=\Im\int_0^\infty\frac{(1+i)x^a}{1+(1+i)x}dx\overset{(1+i)x=u}{=}\Im \frac{1}{(1+i)^a}\int_0^\infty\frac{u^a}{1+u}du$$
$$=-\Im\frac{\pi\csc(a\pi)}{(1+i)^a}=2^{-a/2}\csc(a\pi)\sin(\frac{\pi}{4}a),$$
and your integral is $I''(0).$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

With $\ds{r \equiv -1 + \ic = \root{2}\expo{-\ic\pi/4}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x + 1}^{2} + 1}\,\dd x} =
\left.\partiald[2]{}{\nu}\int_{0}^{\infty}{x^{\nu - 1} \over
\pars{x - r}\pars{x - \overline{r}}}\,\dd x
\,\right\vert_{\ \nu\ = 1^{\large -}}
\\[5mm] = &
\left.\partiald[2]{}{\nu}\int_{0}^{\infty}x^{\nu - 1}\pars{%
{1 \over r - x} - {1 \over \overline{r} - x}}
{1 \over \overline{r} - r}\,\dd x
\,\right\vert_{\ \nu\ = 1^{\large -}}
\\[5mm] = &\
-\left.\partiald[2]{}{\nu}\Im\int_{0}^{\infty}
{x^{\nu - 1} \over r - x}\,\dd x
\,\right\vert_{\ \nu\ = 1^{\large -}}
\\[5mm] = &\
-\partiald[2]{}{\nu}\Im\pars{{1 \over r}\int_{0}^{\infty}
{x^{\pars{\color{red}{\nu} - 1}} \over 1 - x/r}\,\dd x}
_{\ \nu\ = 1^{\large -}}
\end{align}

Note that
$\ds{{1 \over 1 - x/r} = 
\sum_{k = 0}^{\infty}\pars{x \over r}^{k} =
\sum_{k = 0}^{\infty}\color{red}{\expo{\ic\pi k}\Gamma\pars{1 + k} \over r^{k}}{\pars{-x}^{k} \over k!}}$

With Ramanujan's Master Theorem;
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x + 1}^{2} + 1}\,\dd x}
\\[5mm] = &
-\partiald[2]{}{\nu}\Im\bracks{{1 \over r}\,
\Gamma\pars{\nu}\,{\expo{-\ic \pi\nu}\,\,\Gamma\pars{1 - \nu} \over r^{-\nu}}}_{\ \nu\ = 1^{\large -}}
\\[5mm] = &\
\partiald[2]{}{\nu}\Im\braces{\pars{-r}^{\nu - 1}\,
\bracks{\Gamma\pars{\nu}\Gamma\pars{1 - \nu}}}
_{\ \nu\ = 1}
\\[5mm] = &\
\partiald[2]{}{\nu}\Im\bracks{%
\pars{1 - \ic}^{\nu - 1}\,
{\pi \over \sin\pars{\pi\nu}}}_{\ \nu\ = 1}
\\[5mm] = &\
\pi\partiald[2]{}{\nu}\bracks{2^{\nu/2 - 1/2}\,\,\sin\pars{\bracks{1 - \nu}\,{\pi \over 4}}
\csc\pars{\pi\nu}}_{\ \nu\ = 1}
\\[5mm] = &\
\bbx{{5\pi^{3} \over 64} + {\pi \over 16}\ln^{2}\pars{2}}
\approx 2.5167
\\ &
\end{align}
A: 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}
A: It is not bad if you write
$$(x+1)^2+1=(x-a)(x-b)$$
$$\frac 1{(x+1)^2+1}=\frac 1{a-b}\left(\frac 1{x-a}-\frac 1{x-b} \right)$$ making that you face two integrals
$$I_c=\int \frac {\log^2(x)}{x-c}\,dx=\log ^2(x) \log (x-c)-2\int \frac{ \log (x) \log (x-c)}{x}\,dx$$
$$I_c=-2 \text{Li}_3\left(\frac{x}{c}\right)+2 \log (x)
   \text{Li}_2\left(\frac{x}{c}\right)+\log ^2(x) \log \left(1-\frac{x}{c}\right)$$
