How to evaluate $\int_0^{\infty} \frac{2(x-\tanh x)}{\sinh x \tanh^2x} \ \text dx$ $$
\int_0^{\infty}\frac{2\left[\,x - \tanh\left(\,x\,\right)\,\right]}
{\sinh\left(\,x\,\right)\tanh^{2}\left(\,x\,\right)}\,\mathrm{d}x
$$
I know that the answer is $\pi^{2}/4$.
I tried to change $\sinh\left(\,x\,\right)$ and $\tanh\left(\,x\,\right)$ to series but it's hard.  I hope to give me full answers with different methods.
 A: Note that:
$$\int\frac{\tanh(x)}{\sinh(x)\tanh^2(x)}~\mathrm dx=-\operatorname{csch}(x)+c$$
$$\begin{align}\int\frac x{\sinh(x)\tanh^2(x)}~\mathrm dx&=2\int\frac{(u+u^{-1})^2}{(u-u^{-1})^3}\frac{\ln(u)}u~\mathrm du\\&=2\int\frac{(u^2+1)^2}{(u^2-1)^3}\ln(u)~\mathrm du\end{align}$$
By PFD,
$$\frac{32}{(x^2-1)^3}=\frac2{(x-1)^3}-\frac3{(x-1)^2}+\frac3{x-1}-\frac3{x+1}-\frac3{(x+1)^2}-\frac2{(x+1)^3}$$
Each may then be handled through a combination of integration by parts, letting $t=u\pm1$, and polylogarithms.  For example,
$$\begin{align}\int\frac{(u^2+1)^2}{(u-1)^3}\ln(u)~\mathrm du&=\int\frac{t^4+4t^3+8t^2+8t+4}{t^3}\ln(1+t)~\mathrm dt\\&=\int t\ln(1+t)+4\ln(1+t)+8\frac{\ln(1+t)}t+8\frac{\ln(1+t)}{t^2}+4\frac{\ln(1+t)}{t^3}~\mathrm dt\\&=\frac12t^2\ln(1+t)+4t\ln(1+t)-\frac12\ln(1+t)-8\frac{\ln(1+t)}t\\&\phantom{=}-4\frac{\ln(1+t)}{t^2}+4\ln(t)-\frac14t^2-\frac72t-\frac4t-8\operatorname{Li}_2(-t)+C\end{align}$$
(Dear lord, tell me I didn't do that wrong)
We may proceed with all the rest to finally conclude that
$$\int\frac{2(x-\tanh(x))}{\sinh(x)\tanh^2(x)}~\mathrm dx=\operatorname{Li}_2(-e^{-x})-\operatorname{Li}_2(e^{-x})+x\ln(1-e^{-x})-x\ln(1+e^{-x})\\+\frac12\tanh(x/2)-\frac12\coth(x/2)-\frac14x\operatorname{csch}^2(x/2)+2\operatorname{csch}(x)-\frac14x\operatorname{sech}^2(x/2)+C$$
(if this one's wrong, I'm blaming it on WA)
And by substituting in the bounds,

$$\int_0^{+\infty}\frac{2(x-\tanh(x))}{\sinh(x)\tanh^2(x)}~\mathrm dx=\frac{\pi^2}4$$

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}
{2\bracks{x - \tanh\pars{x}} \over \sinh\pars{x}\tanh^{2}\pars{x}}\,\dd x
=
\int_{0}^{\infty}\bracks{%
{2x\cosh^{2}\pars{x} \over \sinh^{3}\pars{x}} - {2 \over x^{2}}}\,\dd x\ +\ 2\
\overbrace{\int_{0}^{\infty}\bracks{%
{1 \over x^{2}} - {\cosh\pars{x} \over \sinh^{2}\pars{x}}}\,\dd x}^{\ds{=\ 0}}
\end{align}
The second integral is trivially evaluated. Indeed,
@Simply Beautiful Art already pointed out in his answer that the second integral involves
$\ds{\dd\bracks{\mrm{csch}\pars{x}}/\dd x = -\cosh\pars{x}/\sinh^{2}\pars{x}}$.

Then,
\begin{align}
&\int_{0}^{\infty}
{2\bracks{x - \tanh\pars{x}} \over \sinh\pars{x}\tanh^{2}\pars{x}}\,\dd x =
\lim_{\epsilon \to 0^{+}}\braces{%
-\int_{x = \epsilon}^{x \to \infty}x\cosh\pars{x}\,
\dd\bracks{1 \over \sinh^{2}\pars{x}}\,\dd x - {2 \over \epsilon}}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\braces{%
{\epsilon\cosh\pars{\epsilon} \over \sinh^{2}\pars{\epsilon}} + \int_{\epsilon}^{\infty}
{1 \over \sinh^{2}\pars{x}}\bracks{\cosh\pars{x} + x\sinh\pars{x}}\,\dd x -
{2 \over \epsilon}}
\\[5mm] = &\
\lim_{\epsilon \to 0^{+}}\bracks{%
{\epsilon\cosh\pars{\epsilon} \over \sinh^{2}\pars{\epsilon}}  + {1 \over \sinh\pars{\epsilon}} + \int_{\epsilon}^{\infty}{x \over \sinh\pars{x}}\,\dd x -
{2 \over \epsilon}} =
\int_{0}^{\infty}{x \over \sinh\pars{x}}\,\dd x
\\[5mm] = &\
2\int_{0}^{\infty}{x\expo{-x} \over 1 - \expo{2x}}\,\dd x =
2\sum_{n = 0}^{\infty}\int_{0}^{\infty}x\expo{-\pars{2n + 1}x}\,\dd x =
2\sum_{n = 0}^{\infty}{1 \over \pars{2n + 1}^{2}} =
2\sum_{n = 0}^{\infty}{1 \over n^{2}} -
2\sum_{n = 0}^{\infty}{1 \over \pars{2n}^{2}}
\\[5mm] = &\
{3 \over 2}\sum_{n = 0}^{\infty}{1 \over n^{2}} =
{3 \over 2}\,{\pi^{2} \over 6} = \bbx{\pi^{2} \over 4}
\end{align}
