Question

How do you integrate \[\int\limits \frac{v-1}{v^2+v+1}dv\]

Answer 1

\[\int\frac{v-1}{v^2+v+1}dv=\int\frac{v+1/2-1/2-1}{v^2+v+1}dv\]

\[=\int \frac{v+1/2}{v^2+v+1}dv-\int\frac{3/2}{v^2+v+1}dv\]

u-sub on the first...complete the square on the 2nd integral

Answer 2

Thank you