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.
 A: It's quite common for such integrals to use algebraic identities in order to solve them, see also here. 
We can use the first one in the link from above, namely:
$$6ab^2=(a+b)^3+(a-b)^3-2a^3\Rightarrow I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)}{x} dx$$
$$=\frac16\int_0^1 \frac{\ln^3(1-x^2)}{x}dx+\frac16\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx-\frac13\int_0^1 \frac{\ln^3(1-x)}{x}dx$$
For the first one let $1-x^2 =t$, for the second one $\frac{1-x}{1+x}=t$ and for the third one $1-x=t$ to get:
$$I=\frac1{12} \int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt-\frac13\int_0^1 \frac{\ln^3 t}{1-t}dt$$
$$=-\frac14\int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt$$
$$=-\frac14\sum_{n=1}^\infty \int_0^1 t^{n-1} \ln^3 t \, dt+\frac13\sum_{n=0}^\infty \int_0^1 t^{2n}\ln^3 t\, dt$$
$$=\frac32\sum_{n=1}^\infty \frac{1}{n^4}-2\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=-\frac38 \zeta(4)=-\frac{\pi^4}{240}$$
A: 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 $I=-\frac{\pi^4}{240}$.
A: 
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.
