Answer
$\frac{1}{3}$$\ln|x^{3}+3x^{2}+9x|$ + C
Work Step by Step
$\int\frac{x^{2}+2x+3}{x^{3}+3x^{2}+9x} dx $
Let $u =x^{3}+3x^{2}+9x$
$\frac{du}{dx}$ = $3x^{2}+6x+9$
$\frac{du}{3x^{2}+6x+9}$ = $dx$
Substitute $u$ and $dx$ into the original equation
$\int\frac{x^{2}+2x+3}{u}\frac{du}{3x^{2}+6x+9}$
= $\int\frac{x^{2}+2x+3}{3x^{2}+6x+9}\frac{1}{u} du$
= $\int\frac{x^{2}+2x+3}{3(x^{2}+2x+3)}\frac{1}{u} du$
= $\int\frac{1}{3}\frac{1}{u} du$
= $\frac{1}{3}$$\int\frac{1}{u} du$
= $\frac{1}{3}$$\ln|u|$ + C
Since $u =x^{3}+3x^{2}+9x$, substituting it back will give you
$\frac{1}{3}$$\ln|x^{3}+3x^{2}+9x|$ + C
