Evaluate $$I=\int_{0}^{1} \frac{\ln(1-x)\ln^2(1+x)\:dx}{x}$$
We have $$\frac{\ln(1-x)}{x}=-\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}$$
Hence $$I=-\sum_{k=1}^{\infty}\left(\frac{1}{k}\int_{0}^{1}x^{k-1}\ln^2(x+1)\:dx\right)\tag{1}$$
Now first lets try to evaluate $$J=\int_{0}^{1}x^{k-1}\ln^2(1+x)\:dx$$ Using Integration By parts and using the fact that $\lim_{x \to 0}x{\ln{x}}=0$
$$J=\frac{\ln^2 2}{k}-\frac{2}{k}\int_{0}^{1}\frac{x^k\ln(1+x)\:dx}{1+x}=J-\frac{2}{k}W \tag{2}$$
Now $$W=\int_{0}^{1}\frac{x^k\ln(1+x)\:dx}{1+x}$$
Writing $$x^k=(1+x-1)^k=(1+x)^k-\binom{k}{1}(1+x)^{k-1}+\binom{k}{2}(1+x)^{k-2}+\cdots+(-1)^k$$
Hence $$W=\sum_{m=0}^{k}(-1)^m\binom{k}{m}\int_{0}^{1}(1+x)^{k-m-1}\ln(1+x)\:dx$$
Using the Fact that when $a\ne -1$: $$\int_{0}^{1}(1+x)^a\ln(1+x)\:dx=\frac{\ln 2 \times 2^{a+1}}{a+1}-\frac{2^{a+1}-1}{(a+1)^2}$$
and when $a=1$ $$\int_{0}^{1}\frac{\ln(1+x)}{1+x}\:dx=\frac{\ln^2 2}{2}$$
Then $$W=\left(\sum_{m=0}^{k-1}(-1)^m\binom{k}{m}\left(\frac{\ln 2 \times 2^{k-m}}{k-m}-\frac{2^{k-m}-1}{(k-m)^2}\right)\right)+\left((-1)^k\frac{\ln^2 2}{2}\right)$$
Using change of variable for the summation as $m=k-r$ we get:
$$W=(-1)^k\left(\sum_{r=1}^{k}(-1)^r\binom{k}{r}\left(\frac{\ln 2 \times 2^{r}}{r}-\frac{2^{r}-1}{(r)^2}\right)\right)+\left((-1)^k\frac{\ln^2 2}{2}\right)$$
Any way to proceed further?
