How can we determine the closed form for $\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}\mathrm dx?$ We know the closed form for $(1)$
$$\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}x\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)-{\zeta(3)\over 2^3}\tag1$$
Proposed:
How can we determine the closed form for $(2)?$

$$\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}\mathrm dx=I\tag2$$

My try:
$y=e^x$
$$\int_{1}^{\infty}{\ln(y-1)\over y+1}\cdot{\mathrm dy\over y}\tag3$$
 A: After the substitution $x=\ln(1+t)$ we have
\begin{equation*}
I = \int_{0}^{\infty}\dfrac{\ln(t)}{(t+1)(t+2)}\, dt \tag{1}
\end{equation*}
which can be evaluated by integrating $\dfrac{\log^{2}(z)}{(z+1)(z+2)}$ along a keyhole contour. As $\log(z)$ we choose the branch given by
\begin{equation*}
\log(z) = \ln|z| +i\arg(z), \quad -\pi<\arg(z) <\pi.
\end{equation*}
We get 
\begin{equation*}
I = \dfrac{1}{2}\ln^{2}(2).
\end{equation*}
Remark. Inspired by $@$FDP I realize that my answer can be simplified by the substitution $t = \frac{2}{s}$ in (1). Then
\begin{gather*}
 I = \int_{0}^{\infty}\dfrac{\ln(2)- \ln(s)}{(s+1)(s+2)}\, ds = \ln(2 ) \int_{0}^{\infty}\left(\dfrac{1}{s+1}-\dfrac{1}{s+2}\right)\, ds -I = \\[2ex]
 \dfrac{1}{2}\ln(2)\left[\ln\left(\dfrac{s+1}{s+2}\right)\right]_{0}^{\infty} = \dfrac{1}{2}\ln^{2}(2).
 \end{gather*}
A: On the path of Mark Viola,
$\displaystyle I=\int_0^\infty \frac{\log(e^x-1)}{e^x+1}\,dx \tag 1$
Perform the change of variable $y=\text{e}^{-x}$,
$\displaystyle I=\int_0^1 \frac{\log(1-x)-\log(x)}{1+x}\,dx\tag 2$
In $(1)$ perform the change of variable $y=\dfrac{1-\text{e}^{-x}}{1+\text{e}^{-x}}$,
$\begin{align} I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1-x}\right)}{1+x}dx\\
&=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1-x)}{1+x}dx\\
&=(\ln 2)^2+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1-x)}{1+x}dx
\end{align}$
Using $(2)$,
$I=(\ln 2)^2-I$
Therefore,
$\boxed{I=\dfrac{(\ln 2)^2}{2}}$
A: Let $I$ be given by the integral
$$I=\int_0^\infty \frac{\log(e^x-1)}{e^x+1}\,dx \tag 1$$
Enforcing the substitution $x\to -\log(x)$ in $(1)$ we find
$$\begin{align}
I&=\int_0^1 \frac{\log(1-x)-\log(x)}{1+x}\,dx\\\\
&=\int_0^1 \frac{\log(1-x)}{1+x}\,dx-\int_0^1 \frac{\log(x)}{1+x}\,dx\tag2
\end{align}$$
We enforce the substitution $x\to 2x-1$ in the first integral on the right-hand side of $(2)$ to reveal
$$\begin{align}
\int_0^1 \frac{\log(1-x)}{1+x}\,dx&=\int_{1/2}^1 \frac{\log(2)+\log(1-x)}{x}\,dx\\\\
&=\log^2(2)-\text{Li}_2(1)+\text{Li}_2(1/2)\\\\
&=\log^2(2)-\frac{\pi^2}{6}+\frac{\pi^2}{12}-\frac12\log^2(2)\\\\
&=\frac12\log^2(2)-\frac{\pi^2}{12}\tag 3
\end{align}$$
We integrate by parts the second integral on the right-hand side of $(2)$ with $u=\log(x)$ and $v=\log(1+x)$ to reveal
$$\int_0^1 \frac{\log(x)}{1+x}\,dx=-\int_0^1 \frac{\log(1+x)}{x}\,dx\tag 4$$
Then letting $x\to -1$ in $(4)$, we find 
$$\begin{align}
\int_0^1 \frac{\log(x)}{1+x}\,dx&=-\int_0^{-1} \frac{\log(1-x)}{x}\,dx\\\\
&=\text{Li}_2(-1)\\\\
&=-\frac{\pi^2}{12}\tag 5
\end{align}$$
Finally, substituting $(3)$ and $(5)$ into $(2)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{I=\frac12\log^2(2)}$$
