\frac{-1}{3}\ln |u-1|+\frac{1}{6}\int \frac{2u+1}{1+u+u^2}\text{du}-\frac{1}{2}\int \frac{1}{1+u+u^2}\text{du}=\frac{-1}{3}\ln |u-1|+\frac{1}{6}\ln |u^2+u+1|-\frac{\sqrt{3}}{3}\text{arctan}\left[\frac{1+2 u}{\sqrt{3}}\right]+C