<s>[INST] Evaluate $\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx$ 
\begin{equation}
\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx
\end{equation}

My colleague got this problem from his friend but he didn't know the answer so he asked my help. Unfortunately, after hours of tired effort I was unable to crack this integral. I was unable to find a way to evaluate it from online search either. I used to be good at solving this kind of problem but now I feel so embarrassed by my stupidity. I'm stuck and I badly need your help. It's a humbling request to ask people here being so kind to help me out. Thank you.
 [/INST] I answer my own OP instead of improving it as a proof to user @mickep that I did give a try to this problem but I didn't want to post some useless efforts like "I tried substitution $x=\tan y$ then I failed... miserably". I consider putting this kind of effort is a a complete joke as shown in some posts with integration tag.
Okay, here is my try. Following user @Anastasiya-Romanova秀's suggestion, I use the substitution $y=e^{-x}$ and the original integral becomes
\begin{equation}
\int_0^1\frac{y(y-1)-y^2\log y}{(1-y)(1+y^2)\log y}dy=-\int_0^1\frac{y}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy
\end{equation}
Now, we consider the following parametric integral
\begin{equation}
I(a):=\int_0^1\frac{y^{a-1}}{1+y^2}\left(\frac{y}{1-y}+\frac{1}{\log y}\right)dy
\end{equation}
Differentiating $I(a)$ we get
\begin{align}
I'(a)& =\int_0^1\left(\frac{\log y}{1-y}+1\right)\frac{y^{a-1}}{1+y^2}dy\\
&=\int_0^1\frac{y^{a}\log y}{1-y^4}dy+\int_0^1\frac{y^{a+1}\log y}{1-y^4}dy+\int_0^1\frac{y^{a-1}}{1+y^2}dy\\
&=\sum_{k=0}^\infty\int_0^1\left(y^{4k+a}\log y+y^{4k+a+1}\log y+(-1)^ky^{2k+a-1}\right)dy\\
&=\sum_{k=0}^\infty\left(-\frac{1}{(4k+a+1)^2}-\frac{1}{(4k+a+2)^2}+\frac{(-1)^k}{2k+a}\right)\\
&=\frac{1}{16}\left(-\psi_1\!\left(\frac{a+1}{4}\right)-\psi_1\!\left(\frac{a+2}{4}\right)+4\psi\!\left(\frac{a+2}{4}\right)-4\psi\!\left(\frac{a}{4}\right)\right)
\end{align}
where I use the following relation
\begin{equation}
\sum_{k=0}^\infty\frac{(-1)^k}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\!\left(\psi_m\left(\frac{z}{2}\right)-\psi_m\!\left(\frac{z+1}{2}\right)\right)
\end{equation}
Hence
\begin{equation}
I(a)=\int_{\infty}^aI'(a)\ da=-\frac{1}{4}\psi\!\left(\frac{a+1}{4}\right)-\frac{1}{4}\psi\!\left(\frac{a+2}{4}\right)+\log\Gamma\!\left(\frac{a+2}{4}\right)-\log\Gamma\!\left(\frac{a}{4}\right)
\end{equation}
Hence our considered problem is $\color{red}{-I(2)}$ which confirms user @ClaudeLeibovici's  result. 
\begin{align}
\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx&=\frac{1}{4}\psi\!\left(\frac{3}{4}\right)+\frac{1}{4}\psi\!\left(1\right)+\log\Gamma\!\left(\frac{1}{2}\right)\\[10pt]
&=\frac{1}{8}\left(\pi+\log\left(\frac{\pi ^4}{64}\right)-4\gamma\right)
\end{align}
where I use special value of the digamma function
\begin{equation}
\psi\!\left(\frac{1}{4}\right)=-\frac{\pi}{2}-3\log2-\gamma
\end{equation}
and the reflection formula
\begin{equation}
\psi\!\left(1-x\right)-\psi\!\left(x\right)=\pi\cot\pi x
\end{equation}
Done!
</s>