We first notice that
\begin{align*}
\int_{0}^{\infty}\frac{\operatorname{gd}(x)}{e^x-1}\,dx
&= \int_{0}^{\infty} \frac{1}{e^x-1} \left( \int_{0}^{x} \frac{dy}{\cosh y} \right) \, dx \\
&= \int_{0}^{\infty} \frac{1}{\cosh y} \left( \int_{y}^{\infty} \frac{dx}{e^x - 1} \right) \,d y \\
&= -2 \int_{0}^{\infty} \frac{\log(1 - e^{-y})}{e^y + e^{-y}} \, dy \\
&=-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta \tag{$e^{-x}=\tan\theta$}.
\end{align*}
The last integral is our starting point. We introduce two tricks to evaluate this.
Step 1. Notice that $\tan(\frac{\pi}{4}-\theta)=\frac{1-\tan\theta}{1+\tan\theta}$. So by the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, it follows that
$$ \int_{0}^{\frac{\pi}{4}}\log(1+\tan\theta)\,d\theta
= \int_{0}^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)\,d\theta $$
and hence both integrals have the common value $\frac{\pi}{8}\log 2$. Applying the same idea to our integral, it then follows that
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta
&= -2\int_{0}^{\frac{\pi}{4}}\log\left(\frac{2\tan\theta}{1+\tan\theta}\right)\,d\theta \\
&= -2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta - \frac{\pi}{4}\log 2.
\end{align*}
Step 2. In order to compute the last integral, we notice that for $\theta\in\mathbb{R}$ with $\cos\theta\neq0$, we have
\begin{align*}
-\log\left|\tan\theta\right|
&= \log\left|\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right|
= \operatorname{Re} \log\left(\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right) \\
&= \operatorname{Re}\left( \sum_{n=1}^{\infty} \frac{1+(-1)^n}{n} e^{2in\theta} \right) \\
&= \sum_{k=0}^{\infty} \frac{2}{2k+1}\cos(4k+2)\theta.
\end{align*}
So by term-wise integration, we obtain
\begin{align*}
-2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta
&= \sum_{k=0}^{\infty} \frac{4}{2k+1} \int_{0}^{\frac{\pi}{4}} \cos(4k+2)\theta \, d\theta \\
&= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}
= 2K,
\end{align*}
where $K$ is the Catalan's constant.
The two answers differs by a constant, as shown below. Let $u= \frac {x-1}{x+1}\tan\frac a2 $, $v=\frac {\sin a}{x+\cos a}$ and evaluate
$$\frac{ u + v}{1-uv}= \tan\frac a2 \cdot \frac{(x+\cos a)(x-1) + 2\cos^2 \frac a2 (x+1)}{(x+\cos a)(x+1)-2\sin^2 \frac a2 (x-1)}=\tan\frac a2$$
Then, utilize the identity $\arctan \frac{ u + v}{1-uv}
=\arctan u + \arctan v $ to get
$$\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right)+\arctan\left(\frac {\sin a}{x+\cos a}\right)=\arctan\left(\tan\frac a2\right) = \frac a2$$
Also, note that
$$\arctan\left(\frac {\sin a}{x+\cos a}\right) = \frac\pi2 -
\arctan\left(\frac {x+\cos a}{\sin a}\right) $$
Thus, the difference of the two is
$$ \csc a\arctan\left(\frac {(x-1)\tan\frac a2}{x+1}\right) -\csc a\arctan\left(\frac {x+\cos a}{\sin a}\right) = \frac12(a-\pi)\csc a$$
which is a constant, as expected.
Best Answer
It's
$$\int_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx= -2\sum\limits_{k=1}^\infty \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$$
which is $\enspace\approx -0.0848704554500727311… $ .
Already $\enspace\displaystyle -2\sum\limits_{k=1}^{10} \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}\enspace$ gives a good approach.
Note: A closed form for such or comparable series is not known to me.
Proof:
$\displaystyle \int_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx=$
$\displaystyle =\int_0^1\lim\limits_{h\to 0}\frac{((1-x+x^2)^h-1)((1+x-x^2)^h-1)}{h^2x}dx$
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\int_0^1\frac{((1-x+x^2)^h-1)((1+x-x^2)^h-1)}{x}dx$
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\left(\int_0^1\left(\frac{(1-(x-x^2)^2)^h-1}{x}-\frac{(1-x+x^2)^h-1}{x}-\frac{(1-x+x^2)^h-1}{x}\right)dx\right) $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\int_0^1\left(\sum\limits_{k=1}^\infty \binom h k \left(x^{k-1}(-x(1-x)^2)^k -x^{k-1}(-1+x)^k -x^{k-1}(1-x)^k\right) \right) $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \binom h k \int_0^1\left(x^{k-1}(-x(1-x)^2)^k -x^{k-1}(-1+x)^k -x^{k-1}(1-x)^k\right)dx $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \binom h k \left((-1)^k\frac{(2k-1)!(2k)!}{(4k)!} -(1+(-1)^k)\frac{(k-1)!k!}{(2k)!}\right) $
$\displaystyle =-\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \left((-1)^{k-1}\binom h k + 2\binom h {2k}\right) \frac{(2k-1)!(2k)!}{(4k)!}$
$\displaystyle =-\sum\limits_{k=1}^\infty \frac{(2k-1)!(2k)!}{(4k)!}\lim\limits_{h\to 0}\frac{1}{h^2}\left((-1)^{k-1}\binom h k + 2\binom h {2k}\right)$
$\displaystyle =-\sum\limits_{k=1}^\infty \frac{(2k-1)!(2k)!}{(4k)!}\frac{H_{2k-1}-H_{k-1}}{k}= -2\sum\limits_{k=1}^\infty \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$
Additional comment:
$$\int_0^1\frac{\log(1-z(x-x^2))\log(1+z(x-x^2))}{x}dx= -2\sum\limits_{k=1}^\infty z^{2k}\frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$$
for $\,z\in\mathbb{C}\,$ and $\,|z|\leq 1\,$ .