How can I solve the integral $ \int {1 \over {x(x+1)(x-2)}}dx$ using partial fractions? $$ \int {1 \over {x(x+1)(x-2)}}dx$$
$$ \int {A \over x}+{B \over x+1}+{C \over x-2}dx $$
I then simplified out and got:
$$1= x^2(A+B+C) +x(C-2B-A) -2A$$
$$A+B+C=0$$
$$C-2B-A=0$$
$$A=-{1 \over 2}$$
However, I'm stuck because I don't know how to solve for B and C now, if I even did the problem correctly.

Well now we know that $B+C = \frac{1}{2}$ and $C-2B = \frac{-1}{2}$. So from that, if we consider this as a system of linear equations, we can deduce that $$B=\frac{1}{3}, C=\frac{1}{6}$$
