Prove that $$-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$$ where ${\rm Li}_2(x)$ is the Poly Logarithm function and $\zeta(s)$ is the Riemann zeta function
Let $$I=-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx $$ Substituting $x=1-y$ gives $$I=-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{1-x}\ dx $$ Adding the above two equations we have $$2I=-\int_{0}^{1}{\rm Li}_2(-x(1-x))\left(\frac{1}{x}+\frac{1}{1-x}\right)\ dx $$ $$2I=-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x(1-x)}\ dx \tag{1}$$ Now we have the integral representation of ${\rm Li}_2(z)$ as $${\rm Li}_2(z)=-\int_{0}^{1} \frac{\log(1-tz)}{t} \ dt$$ For $z=-x(1-x)$ we have $${\rm Li}_2(-x(1-x))=-\int_{0}^{1} \frac{\log(1+tx(1-x))}{t} \ dt\tag{2}$$ So from $(1)$ and $(2)$ $$2I=\int_{0}^{1}\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x(1-x)}\ \frac{\log(1+tx(1-x))}{t} \ dt\ dx $$ Now I am unable to apply integration by parts in the last equation.
Edit I came across this question on MSE click here. I am unable to understand the following equality $$-\frac{1}{2}\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{x}\:dx-\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{1-x}\:dx=-\frac{4}{3}\ln ^3\left(\phi \right)-2\sum _{k=1}^{\infty }\frac{1}{k^3}+2\sum _{k=1}^{\infty }\frac{1}{k^3\:\phi ^{2k}}+4\ln \left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k^2\:\phi ^{2k}}+4\ln ^2\left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k\:\phi ^{2k}}$$ The first integral I am able to evaluate. I am having problem in the second integral. Also how do we get the following? $$2\sum _{k=1}^{\infty }\frac{1}{k^3\:\phi ^{2k}}+4\ln \left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k^2\:\phi ^{2k}}+4\ln ^2\left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k\:\phi ^{2k}}=\frac{8}{5}\zeta \left(3\right)+\frac{4}{3}\ln ^3\left(\phi \right)-\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)+\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)$$