I've been struggling a lot with partial fractions - for instance, with the integral $$\int\frac{x^2 − x + 12}{x^3 + 6x} dx$$ I get that A is 2, but I'm not getting the right answers for B and C. I looked up that they should both be -1, but I can't figure out why.
Does anyone have some tips that might help those of us who are confused with partial fractions? I'm not quite sure what I'd be looking for here, but honestly, nothing could hurt at this point.
Thanks from this hyper-stressed college student!
EDIT: I've been using this website to help me solve these, but it's not showing how to find A, B, and C.
The best thing to do for partial fraction decomposition is to just practice. I’ll outline how to decompose your fraction and provide some practice problems at the end with the answers hidden.
We first rewrite it as$$\frac {x^2-x+12}{x(x^2+6)}=\frac A{x}+\frac {Bx+C}{x^2+6}\quad\implies\quad x^2-x+12=A(x^2+6)+x(Bx+C)$$Now comes the substitution part. We can set $x=0$ to clear one of the terms. So$$12=6A\quad\implies\quad A=\color{blue} 2$$Substitute this back into the problem and simplify to get$$x^2-x+12=\color{blue}2x^2+12+Bx^2+Cx\quad\implies\quad x^2-x+12=x^2(B+2)+Cx+12$$Comparing the coefficients of both sides, we see that $B=C=-1$ which is just what your book says.
Practice: As promised, here are some practice problems.
If you need help, ask in the comment section.
Extra: A nice way to decompose a partial fraction with a linear term $x-a$ multiplied against a polynomial is to differentiate and evaluate at $a$. If we let$$F(x)=(x-a)\psi(x)\quad\implies\quad F'(a)=\psi(a)$$Then we have$$\frac {\varphi(x)}{F(x)}=\frac {A}{x-a}+\frac {\chi(x)}{\psi(x)}\quad\implies\quad \varphi(x)=A\psi(x)+\chi(x)(x-a)$$So$$A=\frac {\varphi(a)}{F'(a)}$$