Integrating $\int_0^1 \frac {\log(1-x)\log^2(1+x)}x \mathrm{d}x$
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 $I=-\frac{\pi^4}{240}$.
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: $$I=\frac1{12} \int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt-\frac13\int_0^1 \frac{\ln^3 t}{1-t}dt$$ $$=-\frac14\int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt$$ $$=-\frac14\sum_{n=1}^\infty \int_0^1 t^{n-1} \ln^3 t \, dt+\frac13\sum_{n=0}^\infty \int_0^1 t^{2n}\ln^3 t\, dt$$ $$=\frac32\sum_{n=1}^\infty \frac{1}{n^4}-2\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=-\frac38 \zeta(4)=-\frac{\pi^4}{240}$$