# Prove $\int_{0}^{1}{(x-1)^2\over \sqrt{2^x-1}\ln(2^x-1)}dx={\pi\over \ln^2{2}}$ [duplicate]

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.

## marked as duplicate by Jack D'Aurizio integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 10 '16 at 11:56

• Since when $(x-1)=(x-1)^2$. If that is true than I agree it is a duplicate. The closed form is also different. – gymbvghjkgkjkhgfkl Jun 10 '16 at 13:54
• @JackD'Aurizio: At a superficial look, at least, the two integrals do not seem identical: this one has a $(x-1)^2$, while the other one only a $x-1$. The upper bound here is $1$, while it is $\infty$ there. Could you give a hint regarding why you consider this one to be a duplicate, please? – Alex M. Jun 10 '16 at 19:04