<s>[INST] 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.
 [/INST] You are almots done. As you wrote $$ \frac{x^2-x+12}{x^3+3x}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$$ then $${x^2-x+12}=A(x^2+3)+(Bx+C)x$$Expand the rhs, so $${x^2-x+12}=(A+B)x^2+Cx+3A$$ Identify the coefficients for a given power of $x$; so $$A+B=1$$ $$C=-1$$ $$3A=12$$ So, $A=4$, $B=-3$,$C=-1$
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