Sub $u=\log{(2^x-1)}$. Then $x=\log{(1+e^u)}/\log{2}$, $dx = (1/\log{2}) (du/(1+e^{-u})$. The integral then becomes
$$\begin{align}\frac{1}{\log{2}} \int_{-\infty}^{\infty} \frac{du}{1+e^{-u}} e^{-u/2} \frac{\frac{\log{(1+e^u)}}{\log{2}}-1}{u} = \frac{1}{2\log^2{2}} \int_{-\infty}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{\frac{1}{2\log^2{2}} \int_{-\infty}^{0} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}}_{u\rightarrow -u} \\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{-\frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^{-u})}-\log{2}}{u}}_{\log{(1+e^{-u})} = \log{(1+e^u)}-u}\\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ \end{align}$$
The nasty pieces of the integral cancel, and we are left with
$$ \frac{1}{2\log^2{2}}\int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} = \frac{\pi}{2 \log^2{2}} $$
as correctly conjectured.
I will attempt to simplify the integral. It might take me a while and I might edit this answer a few times, so this is not its final form yet.
Let $D$ be the differential operator.
We know $\displaystyle\int \dfrac{dx}{\sqrt{x^2-1}}=\ln(2(x+\sqrt{x^2-1}))$.
So by using integration by parts we reduce the problem to solving $$\displaystyle\mathcal{A}=\int_1^\infty D\left[\operatorname{arccot}\left(1+\frac{2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))\, dx.$$
We know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ hence by the chain rule for derivatives we get
$$\mathcal{A}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(2(x+\sqrt{x^2-1}))}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Because we know that $D[\mathrm{arccot}(x)] = \dfrac{-1}{1+x^2} $ and also that $ln(2z)=ln(2)+ln(z)$ we can use integration by parts again and arrive at
$$\mathcal{B}=\int_1^\infty \dfrac{D\left[\left(\frac{-2\,\pi}{\operatorname{arcoth}x\,-\,\operatorname{arccsc}x}\right)\right] \ln(x+\sqrt{x^2-1})}{1+\left(1+\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}\right)^2}\,dx.$$
Now we can turn this into an infinite sum because we set $U=\dfrac{2\, \pi}{\mathrm{arcoth}(x) - \mathrm{arccsc}(x)}$ and use the Taylor expansion for $\frac{z}{1+(1+z)^2}$.
In other words we substitute $z=U$.
( Remember that $\int \Sigma = \Sigma \int$ )
Finally the core problem is reduced mainly to solving:
$$\displaystyle\mathcal{C_n}=\int_1^\infty \frac{\ln(x+\sqrt{x^2-1})}{(\mathrm{arcoth}(x)-\mathrm{arccsc}(x))^n}\, dx.$$
by induction we get the need to solve for $C_1$ and then are able to get the others.
At this point, I must admit that I have ignored convergeance issues.
Those issues can be solved by taking limits.
For instance $C_1$ does not actually converge by itself.
For all clarity the problem is not resolved.
In fact it might require a new bounty. Still thinking about it ...
(in progress)
Best Answer
I think the most ecological approach to the problem is as follows:
Denote $x=\sin\varphi$ and recall that $\arctan x=\frac{1}{2i}\ln\frac{1+ix}{1-ix}$. One then obtains the integral $$\frac{1}{2i}\int_{-\pi/2}^{\pi/2}\ln\frac{4\sqrt{21}+i(11-6\sin\varphi)}{4\sqrt{21}-i(11-6\sin\varphi)}d\varphi=\frac{1}{4i}\int_{0}^{2\pi}\ln\frac{4\sqrt{21}+i(11-6\cos\varphi)}{4\sqrt{21}-i(11-6\cos\varphi)}d\varphi,\tag{1}$$ where at the last step we first used the symmetry of sine function to extend the integration to interval $[-\pi,\pi]$, and then made use of periodicity to shift the integration interval and to replace $\sin$ by $\cos$.
There is a well-known integral (see, for example, here) $$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos\varphi\right)d\varphi=\begin{cases} 0, &\text{for}\; |r|<1,\\ 2\pi\ln r^2, &\text{for}\; |r|>1. \end{cases}\tag{2}$$ Obviously, our integral above is a difference of two integrals of this type. Some care should be taken over multivaluedness of logarithms. This can be handled by saying that the arguments of $4\sqrt{21}\pm i(11-6\cos\varphi)$ belong to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.
Now we have \begin{align} 4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right) \end{align} with $$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}.$$ Since $|r_{\pm}|<1$, the integral (1) reduces to $$\frac{1}{4i}\cdot2\pi\cdot \ln\frac{A_+}{A_-}=\frac{\pi^2}{6}.$$