Finding the closed form of $\int_0^1 \frac{(1-x+x\log x)\operatorname{Li}_3(x)}{x(x-1) \log x} \ dx$ $\def\Li{{\rm{Li}}}$Here I have a question I just received, and  still trying to find a proper starting point
$$\int_0^1 \frac{(1-x+x\log x)\Li_3(x)}{x(x-1) \log x} \ dx$$
What starting point would you propose?

$\def\Li{{\rm{Li}}}$Thanks to @David H's comment I got that 
$$\int_0^1 \frac{(1-x+x\log x)\Li_3(x)}{x(x-1) \log x} \ dx=\frac{5}{4}\zeta(4)-\gamma \zeta(3)+\zeta'(3)$$ that is proving to be numerically correct.
Here I have a question I just received, and  still trying to find a proper starting point
An alternative way
$$\int_{0}^{1}\sum_{n=1}^{\infty}\left(\frac{x^n}{n^3(x-1)}-\frac{x^{n-1}}{n^3\ln{(x)}}\right) \ dx $$
$$=-\sum_{n=1}^{\infty}\frac{1}{n^3}\int_{0}^{1}\left(\frac{x^{n-1}}{\ln{(x)}}+\frac{x^n}{1-x}\right) \ dx \tag{*}$$
$$=-\sum_{n=1}^{\infty}\frac{\log(n)-\psi(n+1)}{n^3}$$
$$=\frac{5}{4}\zeta(4)-\gamma \zeta(3)+\zeta'(3)$$
where in $(*)$ I use the Proposition $2.16.$ from this paper.
Q.E.D.
