## anonymous 4 years ago How do you integrate $\int\limits \frac{v-1}{v^2+v+1}dv$

1. Zarkon

$\int\frac{v-1}{v^2+v+1}dv=\int\frac{v+1/2-1/2-1}{v^2+v+1}dv$ $=\int \frac{v+1/2}{v^2+v+1}dv-\int\frac{3/2}{v^2+v+1}dv$ u-sub on the first...complete the square on the 2nd integral

2. anonymous

Thank you