General formula for $\int^1_0 x^\alpha \log(1-x)\operatorname{Li}_2 (x)\, \mathrm dx$

This is immediately revealed by combining the Cauchy product $$ \log(1-x)\operatorname{Li}_2(x)=3\sum_{n=1}^{\infty} \frac{x^n}{n^3}-2\sum_{n=1}^{\infty} x^n\frac{H_n}{n^2}- \sum_{n=1}^{\infty} x^n \frac{H_n^{(2)}}{n} ,\ |x|\le1 \land x\neq1$$ and well-known harmonic series in the mathematical literature. You might find useful results in (Almost) Impossible Integrals, Sums, and Series and in the paper Euler sum involving tail by Moti Levy and C.I. Valean, here.

More precisely, we need slightly adjusted forms (or we let them the way they are and rearrange the series result obtained after integration) of $$\sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+n+1)}=\frac{H_n^2+H_n^{(2)}}{2n}$$ and $$\sum_{k=1}^{\infty} \frac{H_k^{(2)}}{(k+1)(k+n+1)}= \zeta(2)\frac{H_n}{n}-\frac{1}{n}\sum_{i=1}^n\frac{H_i}{i^2},$$ as they appear in the generalization from Sect. 4.16, page 289 in (Almost) Impossible Integrals, Sums, and Series.