Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach 
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$
multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have
$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$
For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.
What I tried is subbing $1-2x=y$ which gives
$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$
$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$
I think I made it more complicated. Any help would be appriciated.

$$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$
$$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$
