How can I solve the integral $ \int {1 \over {x(x+1)(x-2)}}dx$ using partial fractions?

Question

$$ \int {1 \over {x(x+1)(x-2)}}dx$$ $$ \int {A \over x}+{B \over x+1}+{C \over x-2}dx $$

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.

Answer 1

You have $$B+C=\frac 12\tag 1$$ $$-2B+C=-\frac 12\tag 2$$ Now $(1)-(2)$ gives you $3B=1$, i.e. $B=\frac 13$ so you can get $C$ from $(1)$.

Answer 2

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 $$B=\frac{1}{3}, C=\frac{1}{6}$$

Answer 3

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?