Solve partial integral $∫ x^2 \ln \left(\frac{1 - x}{1 + x} \right)dx$ Hello there this is my first post on this stack exchange community. I joined this community because i'm on my 2nd year of studying software engineering and as one of this year subjects is MA202, where we learn all about the joys of integrals. 
So this is my problem, in the title, i know it's a partial integration because we have a polynomial and a logarithmic function and that i need to use  :
$\ln ((1 - x) / (1 + x)) = u $ and $x^2 = dv$ after solving these i get stuck on the second step where i get to a point where i need to do $∫(1+x)/(1-x)$ which, when solved, returns a logarithmic function which essentially puts me back in the same position i started two steps ago. Can someone give me a hint of what do i need to do or where or what am i doing wrong ? 


$$\int x^2 \ln \left(\frac{1 - x}{1 + x} \right)dx$$

Hint:
By parts $f=\left(\frac{1-x}{1+x}\right)$ and $g=\frac{x^3}{3}$
$$=\frac 1 3 x^3 \ln\left(\frac{1-x}{1+x}\right)-\frac 2 3\int\frac{x^3}{x^2-1}$$
Do long division:
$$=\frac 1 3 x^3 \ln\left(\frac{1-x}{1+x}\right)-\frac 2 3\int x dx-\frac 2 3\int\frac{dx}{2(x-1)}-\frac 2 3\int\frac{dx}{2(x+1)}=\dots$$
