TITLE: Proving that $\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$
QUESTION [11 upvotes]: Could we possibly prove this result without using the polylogarithm? I know how to do it
by   polylogarithm means, but I want a different way. Is that possible?
$$\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$$

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x
     ={35 \over 32}\,\zeta\pars{3} + {1 \over 24}\,\ln^{3}\pars{2}
     -{5 \over 96}\,\pi^{2}\ln\pars{2}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{%
\int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x}
=\Re\int_{0}^{1}{\ln^{2}\pars{1 - x} \over \ic + x}\,\dd x
=\Re\int_{0}^{1}{\ln^{2}\pars{x} \over \ic + 1 - x}\,\dd x
\\[3mm]&=\Re\int_{0}^{1/\pars{1 + \ic}}
{\ln^{2}\pars{\bracks{1 + \ic}x} \over 1 - x}\,\dd x
=\Re\int_{0}^{\pars{1 - \ic}/2}\ln\pars{1 - x}
\bracks{2\ln\pars{\bracks{1 + \ic}x}\,{1 \over x}}\,\dd x
\\[3mm]&=-2\Re\int_{0}^{\pars{1 - \ic}/2}{{\rm Li}_{1}\pars{x} \over x}\,
\ln\pars{\bracks{1 + \ic}x}\,\dd x
=-2\Re\int_{0}^{\pars{1 - \ic}/2}{\rm Li}_{2}'\pars{x}
\ln\pars{\bracks{1 + \ic}x}\,\dd x
\\[3mm]&=2\Re\int_{0}^{\pars{1 - \ic}/2}{{\rm Li}_{2}\pars{x} \over x}\,\dd x
=2\Re\int_{0}^{\pars{1 - \ic}/2}{\rm Li}_{3}'\pars{x}\,\dd x
\end{align}

$$
\color{#c00000}{%
\int_{0}^{1}{x\ln^{2}\pars{1 - x} \over 1 + x^{2}}\,\dd x}
=2\,\Re{\rm Li}_{3}\pars{1 - \ic \over 2}
$$

With one of the MW formulas in group $\pars{1}$:
  \begin{align}
&\overbrace{{\rm Li}_{3}\pars{\half - {\ic \over 2}}
+{\rm Li}_{3}\pars{\half + {\ic \over 2}}}
^{\ds{2\,\Re{\rm Li}_{3}\pars{1 - \ic \over 2}}}
+{\rm Li}_{3}\pars{1 - {2 \over 1 - \ic}}
\\[3mm]&=\zeta\pars{3} + {1 \over 6}\,\ln^{3}\pars{1 - \ic \over 2}
+ {1 \over 6}\,\pi^{2}\ln\pars{1 - \ic \over 2}
-\half\,\ln^{2}\pars{1 - \ic \over 2}\ln\pars{1 + \ic \over 2}
\end{align}

I trust you can take from here.