Improper integral with a dilogarithm: $\int_{0}^{1}\frac{\operatorname{Li}_{2}(x) \ln^{2}(1-x)}{x}~dx$ 
I would like to evaluate the following improper integral:
  $$\int_{0}^{1}\frac{\operatorname{Li}_{2}(x) \ln^{2}(1-x)}{x}~dx$$

I don't know how to start, can I use integration by parts from the beginning?

My trial was to represent  $\operatorname{Li}_{2}(x)$  as a series: $$\sum_{k=1}^{k=\infty}\frac{x^{k}}{k^2}$$
Then, the integral becomes: $$\sum_{k=1}^{k=\infty}\frac{1}{k^2}\int_{0}^{1}x^{k-1}\ln^2(1-x) dx $$ Then use integration by parts but which $u$ and which $dv$ and what to do after that?

By Euler's Beta function
$$ \int_{0}^{1} x^{k-1}(1-x)^a\,dx = B(k,a+1) = \frac{\Gamma(k)\,\Gamma(a+1)}{\Gamma(k+a+1)} \tag{1}$$
hence
$$ \int_{0}^{1} x^{k-1}\log^2(1-x)\,dx = \frac{d}{da^2}\left.\frac{\Gamma(k)\,\Gamma(a+1)}{\Gamma(k+a+1)}\right|_{a=0} = \frac{H_k^2+H_k^{(2)}}{k}\tag{2}$$
and the wanted integral equals
$$ \sum_{k\geq 1}\frac{H_k^2+H_k^{(2)}}{k^3} = \color{red}{2\,\zeta(2)\,\zeta(3)-\zeta(5)}\approx 2.91768094545122231\tag{3}$$
by standard results on Euler sums.
