A problem about partial fraction

Question

I met a strange problem about partial fraction. It asks me to find the partial fraction of $\frac{(s+1)(s+2)(s+3)}{s(s^2+s+1)}$. I know I should use $s^2+s +\frac{1}{4}$, but what to do next?

Thanks

Answer 1

Hint: Rewrite your equation $\frac{(s+1)(s+2)(s+3)}{s(s^2+s+1)}$ in terms of $A$, $B$ and $C$:

$$\frac{(s+1)(s+2)(s+3)}{s(s^2+s+1)}=\left(\frac{A}{s}+\frac{Bs+C}{s^2+s+1}\right)\left(s+3\right)$$

This means that: $$(s+1)(s+2)=s^2+3s+2 \implies (A+B)s^2+(A+C)s+A=s^2+3s+2$$ Therefore: $$\left(\frac{2}{s}-\frac{s-1}{s^2+s+1}\right)\left(s+3\right)=\frac{2s+6}{s}-\frac{s^2+2s-3}{s^2+s+1}\implies 2+\frac{6}{s}-\frac{s^2+s+1+s-4}{s^2+s+1}$$

You can see that the last term can be simplified into: $$1+\frac{6}{s}-\frac{s-4}{s^2+s+1}$$