Answer
$\frac{1}{3}$$\ln|x^{3}+6x^{2}+5|$ + C
Work Step by Step
$\int\frac{x^{2}+4x}{x^{3}+6x^{2}+5} dx $
Let $u =x^{3}+6x^{2}+5$
$\frac{du}{dx}$ = $3x^{2}+12x$
$\frac{du}{3x^{2}+12x}$ = $dx$
Substitute $u$ and $dx$ into the original equation
$\int\frac{x^{2}+4x}{u}\frac{du}{3x^{2}+12x}$
= $\int\frac{x^{2}+4x}{3x^{2}+12x}\frac{1}{u} du$
= $\int\frac{x^{2}+4x}{3(x^{2}+4x)}\frac{1}{u} du$
= $\int\frac{1}{3}\frac{1}{u} du$
= $\frac{1}{3}$$\int\frac{1}{u} du$
= $\frac{1}{3}$$\ln|u|$ + C
Since $u =x^{3}+6x^{2}+5$, substituting it back will give you
$\frac{1}{3}$$\ln|x^{3}+6x^{2}+5|$ + C
