# How to use partial Fractions on $\frac{1}{x(x^2+1)}$?

So I started with $$\frac A x + \frac B{x^2 + 1}.$$ Do I then multiply and get $1=A(x^2 +1) + B(x)$ ? If so I don't know what to do after this.

• Because of the $x^2+1$ it should be $\frac A x+\frac {B+Cx}{x^2+1}$ – Claude Leibovici Mar 15 '15 at 17:16
• You need to start with $\frac{A}{x} + \frac{Bx+C}{x^2+1}$. – Thomas Andrews Mar 15 '15 at 17:16
• Thanks so if x is 0 I get A=1. I don't understand for B, does it need to be the same x=0 value because B would change depending on what I choose for x correct? – Nicole Mar 15 '15 at 17:18
• Thank you thank you that's where i'm going wrong Thomas! – Nicole Mar 15 '15 at 17:18
• So I come up with 1=A(x^2 + 1) +(B+Cx)(x). I'm still confused how to solve each. I know it is simple and from techniques of linear algebra but I'm not seeing it – Nicole Mar 15 '15 at 17:24

## 2 Answers

Your partial fraction decomposition should be $$\frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{1}{x(x^2+1)}$$ Then multiply by $x(x^2+1)$ $$A(x^2+1)+x(Bx+C)=1$$ Let $x=0$ to give: $$A=1$$ Let $x=1$, then $$(1+1)+(B+C)=1$$ $$2+B+C=1$$ $$B+C=-1$$ Let $x=-1$ $$2-(-B+C)=1$$ $$-(-B+C)=-1$$ $$-B+C=1$$ Adding the two equations together gives $$2C=0$$ so $C=0$ and $B=-1$.

• so for solving B, why pick x=1 I'm not seeing that why not x=2? – Nicole Mar 15 '15 at 17:27
• 2 would work equally as well as 1. However with 1 the algebra is a lot simpler. You could have picked 7 or pi or 53343433 or whatever. The point is you just have to pick SOMETHING. Typically -1,0,1 are easiest. – user223391 Mar 15 '15 at 17:29
• ok got it now! thanks so much! – Nicole Mar 15 '15 at 17:31

$$\frac{1}{x(x^2+1)}=\frac A{x}+\frac {Bx+C}{x^2+1}$$ Since $x^2+1$ is irreducible. I found $A=1,B=-1,C=0$