What about of this integral involving the Gudermannian function $\int_0^\infty \frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx$? The Gudermannian function is related to the exponential function, see for example the MathWorld's article Gudermannian. From this idea I was playing with the integral $$\int_0^\infty e^{-nx}\operatorname{gd}(x)\mathrm dx$$
where $n\geq 1$ are integers, when by summation over all these $n$, I wondered about the integral $$\int_0^\infty\frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx.$$
Using Wolfram Alpha online calculator I know the indefinite integral $\int e^{-nx}\operatorname{gd}(x)\mathrm dx$, and also approximations (see the code, or similars, 10 digits of  int gd(x)/(e^x-1)dx, from x=0 to infinity) for which searching with Wolfram Alpha online calculator for a closed-form of the mentioned approximations, I got the following conjecture.
Claim(?). Seems that $$\int_0^\infty \frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx=2K-\frac{\pi \log 2}{4},\tag{C}$$
being $K$ the Catalan's constant.

Question. Is it possible to get a proof of previous conjecture $(\text{C})$? Many thanks.

 A: 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.
A: As shown by Sangchul Lee, the original integral equals
$$ -2\int_{0}^{1}\frac{\log(1-t)}{1+t^2}\,dt \tag{A}$$
and since $\int_{0}^{1}t^{2n}\log(1-t)\,dt=-\frac{H_{2n+1}}{2n+1}$, the object above can be written as
$$ 2\sum_{n\geq 0}(-1)^n\frac{H_{2n+1}}{2n+1}=2K+2\sum_{n\geq 0}\frac{(-1)^n H_{2n}}{2n+1}.\tag{B} $$
On the other hand $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ implies $\sum_{n\geq 1}H_n x^n = -\frac{\log(1-x)}{1-x}$ and
$$ \sum_{n\geq 1}\frac{H_{n-1}}{n}x^n = \frac{1}{2}\log^2(1-x)\tag{C}$$
for any $x\in\mathbb{C}$ such that $|x|<1$. The RHS of $(C)$ is continuous in a neighbourhood of $x=\pm i$ and by considering the principal determination of the complex logarithm we have
$$ \log(1\pm i) = \frac{1}{2}\log 2 \pm i \frac{\pi}{4}, \tag{D}$$
hence by evaluating $(C)$ at $\pm i$ we recover the wanted closed form for the RHS of $(B)$:
$$ -2\int_{0}^{1}\frac{\log(1-t)}{1+t^2}\,dt = 2K-\frac{\pi}{4}\log 2.\tag{E}$$
A: 
In the spirit of my own answer to another integral containing the Gudermmannian function I figured out another possible approach using the relation $\operatorname{gd}(x)=2\arctan(e^x)-\frac\pi2$ combined with the usage of the Clausen Function $\operatorname{Cl}_2(z)$.

Starting with the given integral we may rewrite it sligthly such that we can enforce the substitution $e^{-x}\mapsto x$ which leaves us with the following
\begin{align*}
\int_0^\infty\frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx&=\int_0^\infty\frac{2\arctan(e^x)-\frac\pi2}{e^x-1}\mathrm dx\\
&=\int_0^\infty\frac{\frac\pi2-2\arctan(e^{-x})}{1-e^{-x}}e^{-x}\mathrm dx\\
&=\int_0^1\frac{\frac\pi2-2\arctan(x)}{1-x}\mathrm dx\\
&=\underbrace{\left[-\left(\frac\pi2-2\arctan(x)\right)\log(1-x)\right]_0^1}_{\to0}-2\int_0^1\frac{\log(1-x)}{1+x^2}\mathrm dx\\
&=-2\int_0^1\frac{\log(1-x)}{1+x^2}\mathrm dx
\end{align*}
From hereon we could continue similiar like Jack D'Aurizio or Sangchul Lee already did. We will exploit a different attempt starting with a very similiar substitution $x\mapsto \tan x$ we conclude via several trigonometric identities that
\begin{align*}
-2\int_0^1\frac{\log(1-x)}{1+x^2}\mathrm dx&=-2\int_0^{\pi/4}\log(1-\tan x)\mathrm dx\\
&=-2\int_0^{\pi/4}\log\left(\frac{\sqrt 2\sin\left(\frac\pi4-x\right)}{\cos x}\right)\mathrm dx\\
&=-\frac\pi4\log(2)-2\int_0^{\pi/4}\log\left(\sin\left(\frac\pi4-x\right)\right)\mathrm dx+2\int_0^{\pi/4}\log(\cos x)\mathrm dx\\
&=-\frac\pi4\log(2)-2\int_0^{\pi/4}\log(\tan x)\mathrm dx
\end{align*}
The aformentioned Clausen Function is capable of expressing the latter integral in a closed-form expression. To be precise we further get
\begin{align*}
-\frac\pi4\log(2)-2\int_0^{\pi/4}\log(\tan x)\mathrm dx&=-\frac\pi4\log(2)-2\left(-\frac12\operatorname{Cl}_2\left(2\frac\pi4\right)-\frac12\operatorname{Cl}_2\left(\pi-2\frac\pi4\right)\right)\\
&=-\frac\pi4\log(2)+2\operatorname{Cl}_2\left(\frac\pi2\right)\\
&=-\frac\pi4\log(2)+2G
\end{align*}
Here $G$ denotes Catalan's Constant which is closely related to the Clausen Function.

$$\therefore~\int_0^\infty\frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx~=~-\frac\pi4\log(2)+2G$$

A: I believe your claim is true, but I don't have a proof and merely used Wolfram Alpha to verify the calculation.  After integrating by parts your integral is equal to 
$$
\int_{0}^{\infty} \left(x-\log{(e^x-1)}\right) \operatorname{sech} x \,dx
$$
Wolfram Alpha gives 
$$
\int_{0}^{\infty} x \operatorname{sech} x \,dx = 2C
$$
and, after making the substitution $u = e^x$, it gives 
$$
2 \int_{1}^{\infty} \frac{\log{\left(u-1\right)}}{1+u^2} du = \frac{1}{4}\pi \log 2
$$ for the second part of the intergal.
