Proving that $\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$ 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?
$$\int_0^1\frac{x \log^2(1-x)}{1+x^2} \ dx = \frac{35}{32}\zeta(3)+\frac{1}{24}\log^3(2) -\frac{5}{96} \pi^2 \log(2)$$
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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.
A: My thoughts on the problem (too long for comment):

Prove that the definite integral $I$ has the stated closed form value:
  $$I:=\int_{0}^{1}\frac{x\log^2{(1-x)}}{1+x^2}\mathrm{d}x;\\
I=\frac{35}{32}\zeta{(3)}-\frac{5}{16}\zeta{(2)}\log{(2)}+\frac{1}{24}\log^3{(2)}.$$

Define the auxiliary function $f(\mu)$ for all $\mu\in\mathbb{R}^+$ by the definite integral,
$$f(\mu):=\int_{0}^{1}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x.$$
Differentiating with respect to $\mu$,
$$\begin{align}
\frac{d^2}{d\mu^2}f(\mu)
&=\int_{0}^{1}\frac{\partial^2}{\partial\mu^2}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x\\
&=\int_{0}^{1}\frac{x(1-x)^{\mu-1}\log^2{(1-x)}}{1+x^2}\mathrm{d}x.
\end{align}$$
Evaluating the second derivative of $f(\mu)$ at $\mu=1$ then yields the integral $I$:
$$\frac{d^2}{d\mu^2}f(\mu)\bigg{|}_{\mu=1}=\int_{0}^{1}\frac{x(1-x)^{0}\log^2{(1-x)}}{1+x^2}\mathrm{d}x=\int_{0}^{1}\frac{x\log^2{(1-x)}}{1+x^2}\mathrm{d}x=:I.$$
The function $f(\mu)$ itself can be represented as an alternating series of beta functions:
$$\begin{align}
f(\mu)&=\int_{0}^{1}\frac{x(1-x)^{\mu-1}}{1+x^2}\mathrm{d}x\\
&=\int_{0}^{1}x(1-x)^{\mu-1}\sum_{n=0}^{\infty}(-1)^nx^{2n}\,\mathrm{d}x\\
&=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{1}x^{2n+1}(1-x)^{\mu-1}\,\mathrm{d}x\\
&=\sum_{n=0}^{\infty}(-1)^n\operatorname{B}{(2n+2,\mu)}\\
&=-\sum_{n=1}^{\infty}(-1)^n\operatorname{B}{(2n,\mu)}.
\end{align}$$
The second derivative of $\operatorname{B}{(2n,\mu)}$ with respect to $\mu$ at $\mu=1$ is (courtesy of WolframAlpha),
$$\frac{\partial^2}{\partial\mu^2}\operatorname{B}{(2n,\mu)}\bigg{|}_{\mu=1}=\frac{(H_{2n})^2}{2n}+\frac{\zeta{(2)}}{2n}-\frac{\psi_1{(2n+1)}}{2n}.$$
Hence,
$$\begin{align}
I
&=\frac{d^2}{d\mu^2}f(\mu)\bigg{|}_{\mu=1}\\
&=-\sum_{n=1}^{\infty}(-1)^n\frac{\partial^2}{\partial\mu^2}\operatorname{B}{(2n,\mu)}\bigg{|}_{\mu=1}\\
&=-\sum_{n=1}^{\infty}(-1)^n\left[\frac{(H_{2n})^2}{2n}+\frac{\zeta{(2)}}{2n}-\frac{\psi_1{(2n+1)}}{2n}\right]\\
&=\sum_{n=1}^{\infty}(-1)^n\left[\frac{\psi_1{(2n+1)}}{2n}-\frac{(H_{2n})^2}{2n}-\frac{\zeta{(2)}}{2n}\right]\\
&=\sum_{n=1}^{\infty}(-1)^n\left[\frac{\psi_1{(2n+1)}}{2n}-\frac{(H_{2n})^2}{2n}\right]-\frac{1}{2}\zeta{(2)}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\\
&=\sum_{n=1}^{\infty}(-1)^n\frac{\psi_1{(2n+1)}}{2n}-\sum_{n=1}^{\infty}(-1)^n\frac{(H_{2n})^2}{2n}+\frac{1}{2}\zeta{(2)}\log{(2)}.
\end{align}$$
...Aaand that's as far as I've gotten.
