Nice little generalization:
$$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n},\quad a=0,1,2,...$$
The point of this post is to save us some calculations in our solutions and here is my proof:
\begin{align} I&=\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}\ dx\overset{x\ \mapsto\ 1-x}{=}\int_0^1 \frac{\ln^ax\ln(2-x)}{1-x}\ dx\\ &=\ln2\int_0^1 \frac{\ln^ax}{1-x}\ dx+\int_0^1 \frac{\ln^ax\ln(1-x/2)}{1-x}\ dx\\ &=\ln2((-1)^aa!\zeta(a+1))-\sum_{n=1}^\infty\frac1{n2^n}\int_0^1\frac{x^n\ln^ax}{1-x}\ dx\\ &=\ln2((-1)^aa!\zeta(a+1))-\sum_{n=1}^\infty\frac{(-1)^aa!\left(\zeta(a+1)-H_n^{(a+1)}\right)}{n2^n}\\ &=\ln2((-1)^aa!\zeta(a+1))-\ln2((-1)^aa!\zeta(a+1))+(-1)^aa!\sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}\\ &=(-1)^aa!\sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n} \end{align}
I came up with such generalization while I was working on some nice harmonic series. Other approaches are appreciated.
UPDATE: Here is another nice rule
$$\int_0^1\frac{\ln^ax\ln(1+x)}{1-x}dx=(-1)^aa!\left(\ln2\zeta(a+1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right)$$
and we can prove it following the same approach above:
\begin{align} \int_0^1\frac{\ln^ax\ln(1+x)}{1-x}dx&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^a\ln^nx}{1-x}dx\\ &=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\left[(-1)^aa!\left(\zeta(a+1)-H_n^{(4)}\right)\right]\\ &=-(-1)^aa!\left(\zeta(a+1)\sum_{n=1}^\infty\frac{(-1)^n}{n}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right)\\ &=-(-1)^aa!\left(-\ln2\zeta(a+1)-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right)\\ &=(-1)^aa!\left(\ln2\zeta(a+1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right) \end{align}
Or
$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}$$
