Possibly the strangest calculus problem?

Rewrite the Integral by expressing $\log(1-x) =-\sum_{m=1}^{\infty}\frac{x^m}{m}$

Therefore , after rearranging terms we get

$\displaystyle = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \int\limits_{0}^{1} \frac{x^{m+n-1}\log^2 x}{mn^2} dx$

$\displaystyle = -\sum_{n=1}^{\infty} \sum_{m=1}^{\infty}\frac{2}{(m+n)^3mn^2}$

$=-2\times \frac{\zeta(6)}{6}$

$=-\frac{\pi^6}{2835}$