Partial Fraction Decomposition in $ \frac{s^3-1}{(s^2+6)^2(s+12)^2} $.

Question

What special considerations do you need to take when decomposing the following fraction and why?

I'm trying to decompose the following:

$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} $$

$$ \frac{s^3-1}{(s^2+6)^2(s+12)^2} = \frac{A}{(s^2+6)^2} + \frac{B}{(s+12)^2} $$

It obviously isn't correct. What to do?

Answer 1

When doing Partial Fractions you need to consider all of the increasing powers and should have (see the handy table in the link):

$$ \dfrac{s^3-1}{(s^2+6)^2(s+12)^2} = \dfrac{As + B}{(s^2+6)} + \dfrac{Cs + D}{(s^2+6)^2}+ \dfrac{E}{(s+12)} + \dfrac{F}{(s+12)^2} $$

You should arrive at:

• $A = \dfrac{758}{140625}$
• $B = \dfrac{6841}{562500}$
• $C = -\dfrac{67}{1875}$
• $D = -\dfrac{167}{3750}$
• $E = -\dfrac{758}{140625}$
• $F = -\dfrac{1729}{22500}$