Calculating $\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$ without using Beta function and Euler sum.

Feynman trick works fine here:

Let

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

$$I(a)=\int_0^1\frac{\ln^2x\ln(1-ax)}{1-x}dx,\quad I(1)=I,\quad I(0)=0$$

$$\Longrightarrow I'(a)=-\int_0^1\frac{x\ln^2x}{(1-x)(1-ax)}dx=2\frac{\text{Li}_3(a)}{a}+2\frac{\text{Li}_3(a)-\zeta(3)}{1-a}$$

$$\therefore I= 2\int_0^1\frac{\text{Li}_3(a)}{a}da+2\underbrace{\int_0^1\frac{\text{Li}_3(a)-\zeta(3)}{1-a}da}_{IBP}$$

$$=2\zeta(4)+2\int_0^1\frac{\ln(1-a)\text{Li}_2(a)}{a}da$$

$$=2\zeta(4)-\text{Li}_2^2(1)$$

$$=2\zeta(4)-\frac{5}{2}\zeta(4)=-\frac12\zeta(4)$$

---

Bonus:

I noticed that this trick works for only even powers of $\ln x$ and by following the same technique we find the generalization:

$$\int_0^1\frac{\ln^qx\ln(1-x)}{1-x}dx=-q!\zeta(q+2)-\frac{q!}{2}\sum_{n=1}^{q-1}(-1)^n\zeta(q-n+1)\zeta(n+1)$$

Some cases:

$$\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}dx=12\zeta^2(3)-18\zeta(6)$$

$$\int_0^1\frac{\ln^6x\ln(1-x)}{1-x}dx=720\zeta(3)\zeta(5)-900\zeta(8)$$

For the case of odd powers of $\ln x$, we will need to use Euler sum or Beta function.

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 $6$.