Integrating $\int_0^1 \frac {\log(1-x)\log^2(1+x)}x \mathrm{d}x$ 
Question: Integrate$$\int\limits_0^1dx\,\frac {\log(1-x)\log^2(1+x)}x=-\frac {\pi^4}{240}$$

I'm curious as to if there is a way to integrate this. I've tried using integration by parts to get$$I=-\frac {\pi^2}6\log^22+2\int\limits_0^1dx\,\frac {\operatorname{Li}_2(x)\log(1+x)}{1+x}$$However, I'm not sure how to continue even further. The polylog in the second integrand seems a bit intimidating and I don't see how the first term even helps.


Show that 
  \begin{eqnarray*}
I_2=\int_0^1 \frac{\ln(1-x) (\ln(1+x))^2}{x} dx = -\frac{\pi^4}{240}.
\end{eqnarray*}

It is easy to show (sub $y=1-x$, geometrically expand the denominator and integrate term by term)
\begin{eqnarray*}
I_0=\int_0^1 \frac{(\ln(1-x))^3 }{x} dx = -6\zeta(4)= -\frac{\pi^4}{15}.
\end{eqnarray*}
Now look at Ali's answer here to see that:
\begin{eqnarray*}
I_1 =\int_0^1 \frac{(\ln(1-x))^2 \ln(1+x)}{x} dx = -\frac{5}{8}\zeta(4)+2 \left( \operatorname{Li_4}\left(\frac12\right)+\frac78\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right).
\end{eqnarray*}
Next use Wolfy to get:
\begin{eqnarray*}
I_3 =\int_0^1 \frac{ (\ln(1+x))^3}{x} dx = 6\zeta(4)-6 \left( \operatorname{Li_4}\left(\frac12\right)+\frac{7}{8}\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right).
\end{eqnarray*}
Finally consider the following linear combination of the above integrals
\begin{eqnarray*}
I_0+3I_1+3I_2+I_3= \int_0^1 \frac{(\ln(1-x^2))^3 }{x} dx = \frac{1}{2} I_0.
\end{eqnarray*}
Now just do the linear algebra & we are done.
