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.

The integral is hard to tackle directly (without using Euler sums), but there is a nice trick (which is literally the same as posed above).

Let $$I = \int_0^1 {\frac{{\ln (1 - x){{\ln }^2}(1 + x)}}{x}dx} \qquad J = \int_0^1 {\frac{{{{\ln }^2}(1 - x)\ln (1 + x)}}{x}dx} $$
We have
$$\begin{aligned} 3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx}  + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx}  &= \int_0^1 {\frac{{{{\ln }^3}(1 - {x^2})}}{x}dx} \\ &= \frac{1}{2}\int_0^1 {\frac{{{{\ln }^3}(1 - u)}}{u}du} \end{aligned}$$
the substitution $x^2 = u$ is used.
Hence $$\tag{1}3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx}   = \frac{{{\pi ^4}}}{{30}}$$
On the other hand,
$$\begin{aligned}\int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx}  - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx}  &= \int_0^1 {\frac{{{{\ln }^3}(\frac{{1 - x}}{{1 + x}})}}{x}dx} \\ &= \int_0^1 {\frac{{2{{\ln }^3}u}}{{(1 - u)(1 + u)}}du} \\ &= \int_0^1 {\frac{{{{\ln }^3}u}}{{1 - u}}du}  + \int_0^1 {\frac{{{{\ln }^3}u}}{{1 + u}}du} 
\end{aligned}$$
the substitution $u=\frac{1-x}{1+x}$ is used.
Giving $$\tag{2} - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx}  =  - \frac{{7{\pi ^4}}}{{120}} $$
Adding $(1)$ and $(2)$ together gives $I=-\frac{\pi^4}{240}$.
