TITLE: A conjectured closed form of $\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$
QUESTION [52 upvotes]: Consider the following integral:
$$\mathcal{I}=\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.$$
I tried to evaluate $\mathcal{I}$ in a closed form (both manually and using Mathematica), but without success. 
However, if WolframAlpha is provided with a numerical approximation $\,\mathcal{I}\approx 3.2694067500684...$, it returns a possible closed form: 
$$\mathcal{I}\stackrel?=\frac\pi{2\,\ln^2 2}.$$
Further numeric caclulations show that this value is correct up to at least $10^3$ decimal digits. So, I conjecture that this is the exact value of $\mathcal{I}$.

Question: Is this conjecture correct?

REPLY [23 votes]: Substitute $(2^x-1) = t^2$ to get,  
$ \text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t $  
Substitute $t \mapsto \dfrac{1}{t} $  
$ \implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}  $  
$ \implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2} $  
$ \implies \text{I} = \dfrac{\pi}{2 \ln^2 2} $