TITLE: Calculating $\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$ without using Beta function and Euler sum.
QUESTION [7 upvotes]: Is it possible to show that
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$
without using the Beta function
$$\text{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and the generalized Euler sum
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k),\quad q\ge 2\ ?$$
By integration by parts I found
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx=\color{blue}{\int_0^1\frac{\ln^2(1-x)\ln x}{x}dx}$$
Setting $1-x\to x$ gives the blue integral again.
This integral seems tough under such restrictions. All approaches are welcome.
thanks.

REPLY [2 votes]: If you are patient, you can compute the antiderivative which turns out to be
$$2 \text{Li}_4(1-x)+2 \text{Li}_4\left(\frac{x-1}{x}\right)-2
   \text{Li}_4(x)-\text{Li}_2(x) \log ^2(x)+\text{Li}_2(1-x) \log
   ^2(1-x)+$$ $$\text{Li}_2\left(\frac{x-1}{x}\right) \log
   ^2\left(\frac{1}{x}-1\right)+2 \text{Li}_3(x) \log (x)-$$ $$2 \text{Li}_3(1-x) \log
   (1-x)-2 \text{Li}_3\left(\frac{x-1}{x}\right) \log
   \left(\frac{1}{x}-1\right)+$$ $$\frac{\log ^4(x)}{4}-\log (1-x) \log
   ^3(x)+\frac{1}{2} \log ^2(1-x) \log ^2(x)$$
However, this seems to be almost (strong understatement) impossible when the exponent is $4$ or $6$.