How to prove $\int_{0}^{-1} \frac{\operatorname{Li}_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2} $ $\def\Li{\operatorname{Li}}$
I wonder how to prove:
$$
\int_{0}^{-1} \frac{\Li_2(x)}{(1-x)^2} dx=\frac{\pi^2}{24}-\frac{\ln^2(2)}{2}
$$
I'm not used to polylogarithm, so I don't know how to tackle it. So any help is highly appreciated.

$\def\Li{{\rm{Li}}_2}$Set $x\mapsto -x$ followed by integration by parts, we have
\begin{align}
\int_0^{-1}\frac{\Li(x)}{(1-x)^2}\,dx&=-\int_0^{1}\frac{\Li(-x)}{(1+x)^2}\,dx\qquad\Rightarrow\qquad u=\Li(-x)\,\,\mbox{and}\,\,dv=\frac{1}{(1+x)^2}\\
&=\left.\frac{\Li(-x)}{1+x}\right|_0^{1}+\int_0^{1}\frac{\ln(1+x)}{x(1+x)}\,dx\\
&=-\frac{\pi^2}{24}+\int_0^{1}\frac{\ln(1+x)}{x}\,dx-\int_0^{1}\frac{\ln(1+x)}{1+x}\,dx\\
&=-\frac{\pi^2}{24}-\Li(-x)\bigg|_0^{1}-\frac{1}{2}\ln^2(1+x)\bigg|_0^{1}\\
&=\frac{\pi^2}{24}-\frac{\ln^22}{2}
\end{align}
