I want to solve the integral $$\mathcal{I}=\int_0^1 \frac{\log^2(1-x)}{x}\mathrm{d}x$$ I tried solving it by twice differentiating the Beta function to obtain the logarithm found in $\mathcal I$ and then evaluate at the required arguments. $$B(x,y)=\int_0^1 t^{x-1} (1-t)^{y-1} \mathrm{d}t \\ \implies \frac{\partial^2}{\partial y^2} B(x,y)=\int_0^1 \log^2(1-t) \,t^{x-1} (1-t)^{y-1} \mathrm{d}t\\ \frac{\partial^2B}{\partial y^2} \bigg|_{x=0,\,y=1}=\mathcal I$$ However, the Beta function diverges when one of the arguments is equal to 0. The integral $\mathcal I$ converges to $2\zeta(3)$. Did I do something wrong? If so, please let me know. I am also looking to evaluate a similar integral, $$\int_0^1 \frac{\log^2(1+x)}{x}\mathrm{d}x$$ but I don't even know where to start, so any hints would be greatly appreciated.
It appears this question as already been answered, but I am still wondering why my approach using the Beta function doesn't work.
