Correct $\int_{0}^{1}{x+x^2\over 1-x+x^2}\ln(-\ln x)\mathrm dx=-\gamma-\ln2\ln3?$ An interesting complicate integral
$$\int_{0}^{1}{x+x^2\over 1-x+x^2}\ln(-\ln x)\mathrm dx=-\gamma-\ln2\ln3$$
How can we show this integral is correct?
This is too complicate for me to make an attempt.
Where $\gamma=0.577216...$ 
 A: Hint. A route. One may write, with the change of variable,
$$
t=-\ln x, \quad x=e^{-t},\quad dx=-e^{-t}dt,
$$
$$\begin{align*}
&{\int}_0^1\frac{(x+x^2)\ln(-\ln x)}{1-x+x^2}dx\\\\&=-{\int}_0^\infty \left(1-e^{-t}\right)^2\frac{\ln t}{1+e^{-3t}}e^{-2t}dt\\\\
&=\sum_{n=0}^\infty{\int}_0^\infty (-1)^{n+1}\left(1-e^{-t}\right)^2e^{-(3n+2)t}\ln t\:dt\\\\
&=\sum_{n=0}^\infty  (-1)^{n+1}\left. \partial_{s} \left({\int}_0^\infty t^se^{-(3n+2)t}\left(1-e^{-t}\right)^2\:dt\right)\right|_{s=0}\\\\
&=\left. \partial_{s} \left(\Gamma(s+1) \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+2)^{s+1}}-2\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+3)^{s+1}}+\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+4)^{s+1}}\right)\right)\right|_{s=0}\\\\
&=\left. \partial_{s} \left(\frac{\Gamma(s+1)}{\color{red}{3}^{s+1}} \left(\zeta^-\left(s+1,\frac23\right)-2\zeta^-\left(s+1,\frac33\right)+\zeta^-\left(s+1,\frac43\right)\right)\right)\right|_{s=0}\\\\
&=-\gamma-\ln\color{blue}{2}\ln\color{red}{3}
\end{align*}$$ where
$$
\zeta^-\left(s+1,a\right):=\frac1{\color{blue}{2}^{s+1}}\zeta\left(s+1,\frac {a+1}2\right)-\frac1{\color{blue}{2}^{s+1}}\zeta\left(s+1,\frac {a}2\right)
$$$\displaystyle \zeta(\cdot,\cdot)$ denoting the standard Hurwitz zeta function taking into account its special values.
