$\int_0^1 \frac{\operatorname{Li}_2(x)}{x-2}\mathrm dx=\zeta(3)-\frac{\pi^2}{4}\ln 2$

Question

I'm trying to show that: $$\int_0^1 \frac{\operatorname{Li}_2(x)}{x-2}\mathrm dx=\zeta(3)-\frac{\pi^2}{4}\ln 2$$ , which I got somewhere on internet. Some change of variables didn't help, so I took to IBP(Integration by Parts). $$-\int_0^1 \frac{\operatorname{Li}_2(x)}{2-x}\mathrm dx=\underbrace{\left.\operatorname{Li}_2(x)\log(2-x)\right|_0^1}_{=0}+\int_0^1\frac{\log(1-x)\log(2-x)}{x}\mathrm dx$$ I tried to solve this logarithmic integral using Taylor series and got: $$-\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-1}\log(2-x)\mathrm dx$$ Had it been $\displaystyle\int_0^1 x^{n-1}\log x\,\mathrm dx$, I would write it as $\dfrac{\mathrm d}{\mathrm dn}\displaystyle\int_0^1 x^{n-1}\mathrm dx$, and would solve accordingly, but here it is $\log(2-x)$ in the integrand and applying IBP now overcomplicates the matter. How can I solve this logarithmic integral?