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$\ds{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,{\dd x\over \tanh\pars{\pi x/2}} =
{\phantom{^{2}}\pi^2\over 8} - {1 \over 2}:\ {\large ?}}$.
\begin{align}
&\mbox{Note that}\quad
{1 \over \tanh\pars{\pi x/2}} = {\expo{\pi x} + 1 \over \expo{\pi x} - 1} =
1 + {2 \over \expo{\pi x} - 1}
\end{align}
such that
\begin{align}
&\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,
{\dd x\over \tanh\pars{\pi x/2}} =
\overbrace{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,\dd x}
^{\ds{1 \over 2}}\ +\
2\
\overbrace{\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}
\,{1 \over \expo{\pi x} - 1}\,\dd x}^{\ds{\mbox{Set}\quad x/2\ \mapsto\ x}}
\\[5mm] & =
{1 \over 2} +
{1 \over 2}\int_{0}^{\infty}{x \over
\bracks{x^{2} + \pars{1/2}^{2}}^{2}}
\,{1 \over \expo{2\pi x} - 1}\,\dd x =
{1 \over 2} -
\left.{1 \over 2}\,\partiald{}{z}
\int_{0}^{\infty}{x \over x^{2} + z^{2}}
\,{1 \over \expo{2\pi x} - 1}\,\dd x\,\right\vert_{\ z\ =\ 1/2}
\end{align}
With Binet Second Formula
( see $\ds{\mathbf{\color{#000}{6.3.21}}}$ in A & S Table ),
\begin{align}
&\int_{0}^{\infty}{x \over \pars{1 + x^{2}}^{2}}\,
{\dd x\over \tanh\pars{\pi x/2}} =
{1 \over 2} - {1 \over 2}\,\partiald{}{z}\bracks{\ln\pars{z} - 1/\pars{2z} - \Psi\pars{z} \over 2}_{\ z\ =\ 1/2}
\\[5mm] = &\
{1 \over 2} - {1 \over 2}\bracks{2 - {1 \over 2}\,\Psi\,'\pars{1 \over 2}} =
{1 \over 4}\,\Psi\,'\pars{1 \over 2} - {1 \over 2} =
\bbx{\color{#44f}{{\phantom{^{2}}\pi^{2} \over 8} - {1 \over 2}}} \\ &
\end{align}
Note that $\ds{\Psi\,'\pars{1/2} = 3\pars{\pi^{2}/6} = \pi^{2}/2}$ ( see $\ds{\mathbf{6.4.4}}$ in A & S Table ).