The function $x\mapsto\frac1{x-1}$ is defined on the disconnected set $\Bbb R\setminus\{1\}$. Its primitives on the interval $(1,+\infty)$ are functions of the form $x\mapsto \ln(x-1)+C_1$ where $C_1$ is a constant, while its primitives on $(-\infty,1)$ are functions of the form $x\mapsto \ln(1-x)+C_2$ where $C_2$ is a constant; its primitives one the whole domain are any combination of a pair of primitives on these segemnts, so there are now two independent constants of integration.
The usual convention of writing $\int f=g$ where $g$ is some primitive of $f$, implicitly assuming that the general form of a primitive of $f$ is $g+C$ only works well if $f$ has a connected domain. So in particular it does not work well for $f:x\mapsto\frac1{x-1}$ defined on $\Bbb R\setminus\{1\}$. If you write $\int f=g_1$ where $g_1(x)=\ln(x-1)$, which is defined for $x>1$ only, then you have only given a primitive on part of the domain; if you write $\int f=g_2$ where $g_2(x)=\ln(1-x)$, which is defined for $x<1$ only, then you have only given a primitive on a complementary part of the domain, and if you write $\int f=g_3$ where $g_2(x)=\ln|1-x|$ as many do, which is defined on all of the domain of $f$, then you falsely suggest that one is obliged to choose the constants of integration to be equal on both components of the domain. So there is no really good solution here.
Apparently your CAS has chosen to write $\int f=g_1$ here, but $\int -f=-g_2$. That may seem inconsistent, since $-f$ is surely defined on the same domain as $f$ is, but neither is entirely wrong, they are just part of the truth. And it is hard to design computer algebra software that must make arbirtrary choices in such a way that the end result never seems inconsistent. Combining the two parts in an equality statement now leads to rewriting equation that looks patently valid into something that is nonsensical, as it is comparing expressions that are defined on disjoints sets. But with indefinite integration you can expect to find even worse than this occasionally, namely that equating the primitives of two expressions that are equivalent (and everywhere defined) leads to an equation that is always false, namely one equating two expressions that differ by a constant (obtained by choosing different constants of integration when finding primitives for both expressions).
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.
Best Answer
Equal to $I(1/2)$ where, for real $a>0$, \begin{align}I(a)&=\int_{-\infty}^{\infty}\log\frac{\cosh(x+a)\cosh(x-a)}{\cosh^2 x}\,dx\\&=\int_{-\infty}^{\infty}\int_{0}^{a}\left(\tanh(x+y)-\tanh(x-y)\right)\,dy\,dx\\&=\int_{0}^{a}\left.\log\frac{\cosh(x+y)}{\cosh(x-y)}\right|_{x=-\infty}^{x=\infty}\,dy=4\int_{0}^{a}y\,dy=2a^2.\end{align}