I want to prove the variation from here
$$\int_{0}^{1}{(x-1)^2\over \sqrt{2^x-1}\ln(2^x-1)}dx=\color{blue}{\pi\over \ln^2{2}}\tag1$$
Sub $u=2^x-1\rightarrow du=2^x\ln{2}dx$
$x=1\rightarrow u=1$ and $x=0\rightarrow u=0$
$(x-1)^2={1\over \ln^2{2}}\ln^2\left({1+u\over 2}\right)$
$$I={1\over \ln{2}}\int_{0}^{1}{(x-1)^2\over \sqrt{u}(u+1)\ln{u}}du\tag2$$
$$I={1\over \ln^3{2}}\int_{0}^{1}{\ln^2\left({u+1\over 2}\right)\over \sqrt{u}(u+1)\ln{u}}du\tag3$$
Sub $u=e^z\rightarrow du=e^zdz$
$u=1\rightarrow z=0$ and $u=0\rightarrow z=-\infty$
${e^z+1\over 2}=e^{z/2}\cosh(z/2)$
$$I=-{1\over \ln^3{2}}\int_{0}^{\infty}{\ln^2(e^{z/2}\cosh(z/2))\over ze^{-z/2}(1+e^z)}dz\tag4$$
$e^{-z/2}(1+e^z)=2\cosh(z/2)$
$$I=-{1\over \ln^3{2}}\int_{0}^{\infty}{\ln^2(e^{z/2}\cosh(z/2))\over 2z\cosh(z/2)}dz\tag5$$
I am stuck! Can anyone demonstrate how to prove I in step by step manner please, thank.
I try to understand their proofs From here, but couldn't follow.
