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