<s>[INST] Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$ Prove that (please)

$$\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$$

I've tried using Taylor series and I ended up with
$$-\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{2}{n(m+n+1)^3}$$
I am stuck there and I couldn't continue it using partial fraction to get any familiar sum of series. Could anyone here please help me to prove the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
 [/INST] The answer by user111187 is the cleanest way to approach such problem. I will elaborate on the sums (which he doesnt show). 
$$I = (-)\cdot\int_{0}^{1} \frac{\log(1-x)\log^2(x) dx}{1-x}$$
Let $u = 1-x \implies du = -dx$ 
$$I = -\int_{0}^{1} \frac{\log(u)\log^2(1-u) du}{u}$$
$$\sum_{n=1}^{\infty} H_n u^n = -\frac{\log(1-u)}{1-u}$$
$$\sum_{n=1}^{\infty} \frac{H_nu^{n+1}}{n+1} = \frac{\log^2(1-u)}{2}$$
$$(2)\cdot\sum_{n=1}^{\infty} \frac{H_n\log(u)\cdot u^{n}}{n+1} = \frac{\log^2(1-u)\log(u)}{u}$$
$$(-2)\cdot \sum_{n=1}^{\infty}\frac{H_n}{n+1}\cdot \int_{0}^{1} \log(u) \cdot u^{n} du = \int_{0}^{1} \frac{\log^2(1-u)\log(u)}{u} du = I$$
For the integral in the LHS, we simply use Leibniz's rule considering $\displaystyle J(n) = \int_{0}^{1} u^n du$ then differentiate.
$$(2)\cdot \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} = \int_{0}^{1} \frac{\log^2(1-u)\log(u)}{u} du = I$$
$$H_n = H_{n-1} + \frac{1}{n} \implies H_{n+1} = H_{n} + \frac{1}{n+1} \implies H_n = H_{n+1} - \frac{1}{n+1}$$
$$ = (2)\cdot \sum_{n=1}^{\infty}\frac{H_{n+1}}{(n+1)^3} - (2)\cdot\sum_{n=1}^{\infty} \frac{1}{(n+1)^4}$$
$$ = (2)\cdot \sum_{n=2}^{\infty} \frac{H_{n}}{n^3} - (2)\cdot\sum_{n=2}^{\infty}  \frac{1}{n^4}$$
$$ = (2)\cdot \left( \sum_{n=1}^{\infty} \frac{H_{n}}{n^3} - 1   \right) - (2) \cdot \left( \sum_{n=1}^{\infty}  \frac{1}{n^4} - 1 \right)$$
From Generalised Euler Sum
$$ = (2)\cdot \left( \left(1 + \frac{3}{2} \right)\zeta(4) - \frac{1}{2}\zeta^2(2) - 1   \right) - 2\zeta(4) + 2$$
$$ = 5\zeta(4) - \zeta^2(2) - 2 - 2\zeta(4) + 2 = 3\zeta(4) - \zeta^2(2) = \frac{\pi^4}{180}$$
Finally,
$$I =  \int_{0}^{1} \frac{\log(1-x)\log^2(x) \space dx}{x-1}  = \frac{\pi^4}{180}$$
</s>