Integrals: Partial Fractions $$ \int \frac{x^2-x+12}{x^3+3x} $$
I factored the denominator to get $ x(x^2+3) $. I then seperated the x and the $x^2+3$ into the partials $\frac{A}{x}$ and $\frac{Bx+C}{x^2+3}$.
After combining the two, I came up with $$ \frac{A(x^2+3)+ x(Bx+C)}{x(x^2+3)}$$
This is where I'm stuck. I'm assuming that A would equal 1 because it is where $x^2$ is in the original problem. The same would go for b equaling -1 and $C = 12$. I'm not so sure though.

Note that $x^2-x+12=12$ when $x=0$.
Note also that $A(x^2+3)+x(Bx+C)=3A$ when $x=0$. Thus $A=\frac{12}{3}=4$.
The coefficient of $x^2$ in $x^2-x+12$ is $1$. In  $A(x^2+3)+x(Bx+C)$ it is $A+B$. Since $A=4$, we have $B=-3$.
We leave finding $C$ to you. 
