-\int \frac{u}{u^3-1}\text{du}=-\int \left(\frac{1}{3(u-1)}+\frac{1-u}{3 \left(1+u+u^2\right)}\right)\text{du}=-\int \frac{1}{3(u-1)}\text{du}+\int \frac{u-1}{3 \left(1+u+u^2\right)}\text{du}=\frac{-1}{3}\ln |u-1|+\frac{1}{6}\int \frac{2u-2+1-1}{1+u+u^2}\text{du}