$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$ Someone please help me to solve this integral
$$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$$
I have tried to use the formula
$$ab^2=\frac{(a+b)^3+(a-b)^3+2a^2}{6}$$
to reduce the original integral to three integrals
$$\frac{1}{6}\int_0^1\frac{\ln^3(x+x^2)}{1-x}\,dx+\frac{1}{6}\int_0^1\frac{\ln^3(\frac{x}{1+x})}{1-x}\,dx+\frac{1}{3}\int_0^1\frac{\ln^3x}{1-x}\,dx$$
But I couldn't figure out how to solve these.
Another way is to use series representation of $$\ln(x), \quad \ln(1+x),\quad \frac{1}{1-x}.$$
But I don't know how to solve the multiple sums.
Please help me out.
 A: $
\displaystyle\int_0^1\frac{\ln^2(x)\ln(1+x)}{1-x}dx=\displaystyle\int_0^1\displaystyle\int_0^1\frac{x\ln^2(x)}{(1+xy)(1-x)}dxdy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(\displaystyle\int_0^1\frac{\ln^2(x)}{1-x}-\frac{\ln^2(x)}{1+yx}dx\right)dy$
$=\displaystyle\int_0^1\frac{1}{1+y}\left(2\zeta(3)-\displaystyle\sum_{n=1}^{\infty}(-y)^{n-1} \displaystyle\int_0^1x^{n-1}\ln^2(x)dx\right)dy$
$=2\zeta(3)\ln(2)-2\displaystyle\int_0^1\frac{1}{1+y} \displaystyle\sum_{n=1}^{\infty}\frac{(-y)^{n-1}}{n^3}dy=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y(1+y)}dy$
$=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y}-\frac{\operatorname{Li}_3(-y)\ }{1+y}dy$
$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)-2\left(\ln(1+y)\operatorname{Li}_3(-y)\ \right)_0^1+\displaystyle\int_0^1\frac{\operatorname{Li}_2(-y)\ \ln(1+y)}{y}dy$
$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)+\frac{3\ln(2)\zeta(3)}{2}-2\displaystyle\int_0^1 \operatorname{Li}_2(-y)(-\operatorname{Li}_2^{'}(-y))dy$
$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}-2\left(-\frac{(\operatorname{Li}_2(-y))^2}{2}\right)_0^1=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+(\operatorname{Li}_2(-1))^2$
$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+\left(-\frac{\pi^2}{12}\right)^2$
