## Calculus (3rd Edition)

$\frac{1}{2}\ln (9-2x+3x^2)+c.$
Let $u=9-2x+3x^2$ and hence $du=2(3x-1)dx$, then we have $$\int \frac{ 3x-1}{9-2x+3x^2}dx=\frac{1}{2}\int \frac{ d u}{u}= \frac{1}{2}\ln |u|+c=\frac{1}{2}\ln |9-2x+3x^2|+c.$$