Question

the integral of (x +2) dx/ x^2 + 4x -9

Answer 1

Let u=x^2+4x-9
du=2x+4 dx
(1/2)du=x+2 dx
So your integral becomes:
\[\frac{1}{2}\int\limits \frac{du}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|x^2+4x-9|+C\]