How to evaluate $\int_0^\infty \frac{e^{-x}+x-1}{x(e^{2x}-e^{-2x})}dx$? We are unable to verify this this equality
$$
4\int\limits_0^\infty \frac{e^{-x}+x-1}{x\left(e^{2x}-e^{-2x}\right)}\;\mathrm{d}x=\gamma+\ln\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\;.
$$
where $\gamma$ is Euler's constant which is approximately $0.5772156\dots$, and $\Gamma(n)$ is the gamma function and the particular value for $\Gamma\left(\frac{1}{4}\right)=3.625690\dots$.
 A: Hint. A possible route. 
One may set
$$
f(s):=4\int_0^\infty \frac{e^{-sx}+sx-1}{x(e^{2x}-e^{-2x})}dx, \quad s>0. \tag1
$$ In order to get rid of the $x$ in the denominator, we may differentiate under the integral sign getting
$$
f'(s)=4\int_0^\infty \frac{1-e^{-sx}}{e^{2x}-e^{-2x}}dx, \quad s>0. \tag2
$$ Then expanding the latter integrand and integrating termwise we get
$$
f'(s)=\gamma+2\ln 2+ \psi\left(\frac12+\frac{s}4\right) \tag3
$$ where $\displaystyle \psi : = \Gamma'/\Gamma$ may be obtained by the following series representation,
$$
\psi(u+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k}  
\right), \quad u >-1, \tag4
$$ $\gamma$ being the Euler-Mascheroni constant.
Integrating $(3)$, with the fact that, as $s \to 0$, 
$$f(s) \sim 2s^2\int_0^\infty\!\! \frac{x}{e^{2x}-e^{-2x}}dx \sim \frac{\pi^2}{16}s^2 \tag5$$
we get

$$
4\int_0^\infty \frac{e^{-sx}+sx-1}{x(e^{2x}-e^{-2x})}dx=\gamma s+2s \ln2-2 \ln \pi+4 \log \Gamma\left(\frac12+\frac{s}4\right), \quad s>0, \tag6
$$ 

from which you deduce the value of your initial integral by putting $s:=1$.
A: Another approach. We have $$\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{k}}{k!}=e^{-x}
 $$ hence $$e^{-x}-1+x=\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}}{k!}
 $$ and so $$I=4\int_{0}^{\infty}\frac{e^{-x}-1+x}{x\left(e^{2x}-e^{-2x}\right)}dx=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k!}\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx
 $$ now note that $$ \begin{align} 
\int_{0}^{\infty}\frac{x^{k-1}}{e^{2x}-e^{-2x}}dx= & \int_{0}^{\infty}x^{k-1}\sum_{m\geq0}e^{-\left(2m+1\right)2x}dx \\ = & \sum_{m\geq0}\int_{0}^{\infty}x^{k-1}e^{-\left(2m+1\right)2x}dx
  \\ =
  & \frac{\left(k-1\right)!}{2^{k}}\sum_{m\geq0}\frac{1}{\left(2m+1\right)^{k}}
  \\ =
  & \frac{\left(k-1\right)!}{2^{k}}\left(1-\frac{1}{2^{k}}\right)\zeta\left(k\right)
 \end{align}$$ and so $$I=4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k2^{k}}\zeta\left(k\right)-4\sum_{k\geq2}\frac{\left(-1\right)^{k}}{k4^{k}}\zeta\left(k\right)
 $$ and now we can use the identity, for $x\geq0,
 $ $$\sum_{k\geq2}\frac{\left(-x\right)^{k}}{k}\zeta\left(k\right)=x\gamma+\log\left(\Gamma\left(x+1\right)\right)
 $$ so finally we have $$I=\gamma+\log\left(\frac{16\pi^{2}}{\Gamma\left(\frac{1}{4}\right)^{4}}\right).
 $$
