Partial fraction decomposition of rational functions

Question

How can I start this with Partial fractions?

$$ \frac{10}{(s^2-4)(s^2+4)}+\frac{1}{s^2+4}$$

I was thinking of something like:

$$\frac{As^2+Bs+C}{s^2-4}+\frac{Ds^2+Es+F}{s^2+4}+\frac{Gs^2+Hs+O}{s^2-4} $$

But then it seems to fairly complicated.

Any ideas?

Answer 1

You can first work with $s^2$ as an atomic variable. Rewrite $$ \frac{10}{(s^2-4)(s^2+4)}+\frac{1}{s^2+4}=\frac{s^2+6}{(s^2-4)(s^2+4)}.$$ Then two terms (hence two unknowns), $$\frac A{s^2-4}+\frac B{s^2+4}$$ will suffice. $$A+B=1\\4A-4B=6,$$ $$A=\frac54,B=-\frac14.$$ You can further decompose the quadratic denominators.

Answer 2

$$\frac{10}{(s^2-4)(s^2+4)}+\frac{1}{s^2+4}=\frac{s^2+6}{(s^2-4)(s^2+4)}$$

Remember that $s^2-4=s^2-2^2=(s-2)(s+2)$

Then your partial fractions will be of the form $$\frac{Ax+B}{s^2+4}+\frac{C}{s+2}+\frac{D}{s-2}$$

Answer 3

Note that $s^2-4=(s+2)(s-2)$ so you are looking for $$\frac {10}{(s^2-4)(s^2+4)}=\frac a{s+2}+\frac b{s-2}+\frac {cs+d}{s^2+4}$$to which you add $\cfrac 1{s^2+4}$ to get final term $\cfrac {cs+d+1}{s^2+4}$