How can we show that $(1)$ $$\int_{0}^{1}{x-3x^3+x^5\over 1+x^4+x^8}\ln(-\ln x)\mathrm dx={\pi \ln 2\over 6\sqrt{3}}?\tag1$$
Trying a substitution $$t=-\ln x, x=e^{-t}, \mathrm dx=-e^{-t}\mathrm dt$$
$$\int_{0}^{\infty}{e^{-2t}-3e^{-4t}+e^{-6t}\over 1+e^{-4t}+e^{-8t}}\ln(t)\mathrm dt\tag2$$
$y=e^{-t}$
$$f(y)={y^2-3^4+y^6\over 1+y^4+y^8}\tag3$$
We need to decomposed the fractions to aid the process of trying to prove $(1)$