<s>[INST] Integration of $\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3}\, dx$ How would you integrate:
$$I=\int^{\pi/45}_0 \frac{x^2 \ln (1-x)}{1-x^3} dx$$
My attempt:
$$du=\frac{x^2}{1-x^3}dx ⇒ u=\frac{-1}{3}\ln (1-x^3)$$
$$v= \ln (1-x) ⇒ dv=\frac{-1}{1-x}$$
$$I=\frac{-1}{3}\ln (1-x^3) \ln (1-x)-\int \frac{\ln (1-x^3)}{1-x}\,dx$$
$\ln (1-x^3)=\ln (1-x)+\ln (1+x+x^2)$
$$\int \frac{\ln (1-x^3)}{1-x}\,dx=\int \frac{\ln(1-x)}{1-x}\,dx+\int \frac{\ln (x^2+x+1)}{1-x}\,dx$$
$$I_1=\int \frac{\ln(1-x)}{1-x}\,dx=-\frac{1}{2} \ln^2 (1-x)$$
$$x^2+x+1=\left(x+\frac{1-i\sqrt3}{2}\right)\left(x+\frac{1+i\sqrt3}{2}\right)$$
$$\ln \left(x+\frac{1-i\sqrt3}{2}\right)= \ln (2x+1-i\sqrt 3)- \ln 2=\tan^{-1}-\frac{\sqrt3}{2x+1}-\ln 2$$
$$\ln \left(x+\frac{1+i\sqrt3}{2}\right)= \ln (2x-1+i\sqrt 3)- \ln 2=\tan^{-1}\frac{\sqrt3}{2x-1}-\ln 2$$
$$\int \frac{\ln (x^2+x+1)}{1-x} \,dx=\int \frac{\tan^{-1}-\frac{\sqrt3}{2x+1}}{1-x}\,dx+\int \frac{\tan^{-1}\frac{\sqrt3}{2x-1}}{1-x}\,dx+2 \ln2 \ln (1-x)$$
Can someone proceed?
 [/INST] Hint: A better method
Using the  partial fraction:
$$\frac{{{x}^{2}}}{1-{{x}^{3}}}=\frac{1}{3}\frac{1}{1-x}-\frac{1}{3}\frac{2x+1}{{{x}^{2}}+x+1}=\frac{1}{3}\frac{1}{1-x}-\frac{2}{3}\frac{1}{2x-i\sqrt{3}+1}-\frac{2}{3}\frac{1}{2x+i\sqrt{3}+1}$$ 
The integral becomes 
$$\begin{align}
  & I=\frac{1}{3}\int{\frac{\ln \left( 1-x \right)}{1-x}dx--\frac{2}{3}\int{\frac{\ln \left( 1-x \right)}{2x-i\sqrt{3}+1}dx}-\frac{2}{3}\int{\frac{\ln \left( 1-x \right)}{2x+i\sqrt{3}+1}dx}} \\ 
 & \quad =-\frac{1}{6}{{\ln }^{2}}\left( 1-x \right)-\frac{2}{3}J-\frac{2}{3}K \\ \end{align}$$
Evaluating $J$ and $K$ are similar, for $J$  first set $u=1-x$, to get:
$$J=\int{\frac{\ln \left( u \right)}{2u-3+i\sqrt{3}}du}$$
Now use the formula:
$$\int{\frac{\ln \left( u \right)}{au+b}dx}=\frac{1}{a}\log \left( u \right)\log \left( \frac{au}{b}+1 \right)+\frac{1}{a}L{{i}_{2}}\left( -\frac{au}{b} \right)$$ 
where $L{{i}_{n}}$ is the poly-logarithm function.
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