Partial fraction decomposition of $\frac{1}{x^4-x^2}$

I have been having a debate over whether, when you factor the denominator into $x^2$, $x-1$, and $x+1$, you need a fraction that says $\frac{A}{x}$ and one that has $\frac{B}{x^2}$ or if you only need the fraction with $x^2$ as the denominator. When I worked it out, I was only able to get the correct answer when I didn't do the $\frac{A}{x}$ fraction, but maybe I made such a roundabout mistake that I happened to get the right thing the wrong way. When do you need two fractions and when don't you?

• You should get $A=0$. – Kenny Lau Sep 5 '17 at 14:58
• Assume that you will need $\frac Ax + \frac B{x^2} + \cdots$ and it may turn out that $A = 0$ – Doug M Sep 5 '17 at 15:12

$$\frac{1}{x^4-x^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}+\frac{D}{x^2}$$
$$\frac {1}{x^4-x^2}=\frac {1}{x^2(x^2-1)}=\frac {x^2-(x^2-1)}{x^2(x^2-1)}=\frac 1{x^2-1}-\frac 1{x^2}.$$
And $$\frac 1{x^2-1}=\frac 12 \frac 2{(x-1)(x+1)}=\frac 12 \frac {(x+1)-(x-1)}{(x-1)(x+1)}=\frac 1{2(x-1)}-\frac 1{2(x+1)}.$$