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.

Generally you want to avoid simultaneous equations. So rather than collect coefficients of powers of $x$ as you have done, write it as $1=A(x+1)(x-1)+Bx(x-1) +Cx(x+1)$. Since this is an identity, you can substitute any value of $x$ into this. Therefore substitute values which will make brackets disappear. For example, putting $x=1$ will give you the value of $B$, and putting $x=0$ will give you the value of $A$, and so on. An even quicker way is to use the Cover-Up Rule. Do you know this?
