How do we prove this ? $$\displaystyle\int_0^1\frac{\log(\log(\frac{1}{x}))}{1-x+x^2}dx=\frac{2\pi}{\sqrt{3}}\log(\frac{\sqrt[6]{32\pi^5}}{\Gamma(\frac{1}{6})})$$ My atempte: \begin{align*} \displaystyle\int_0^1\frac{\log(\log(\frac{1}{x}))}{1-x+x^2}dx&=\displaystyle\int_1^{\infty}\frac{\log(\log(t))}{1-t+t^2}dt&(\frac{1}{x}=t)\\ &=\displaystyle\int_0^{\infty}u\frac{e^u}{(e^u-1)^2+e^u}du&(\log(t)=u)\\ &=\displaystyle\int_0^{\infty}u\frac{1}{1+\left(\frac{e^u-1}{e^{\frac{u}{2}}}\right)^2}du\\ &=\displaystyle\sum_{n=0}^{\infty}(-1)^n\displaystyle\int_0^{\infty}u(e^{-nu})(e^u-1)^{2n}du\\ \end{align*}
In the last integration, I had no idea to complete the proof
Any help to prove this
And thanks in advance
