Originally Posted by

**Siron** I'v done it like this:

$\displaystyle \frac{3s}{(s^2+4)\cdot(s^2+1)}=\frac{As+C}{s^2+4}+ \frac{Bs+D}{s^2+1}$

$\displaystyle \Leftrightarrow \frac{(As+C)\cdot (s^2+1)+(Bs+D)\cdot (s^2+4)}{(s^2+4)\cdot(s^2+1)}$

$\displaystyle \Leftrightarrow \frac{As^3+Cs^2+As+C+Bs^3+Ds^2+4Bs+4D}{(s^2+4)(s^2 +1)}$

$\displaystyle \Leftrightarrow \frac{(A+B)s^3+(C+D)s^2+(A+4B)s+(C+D)}{(s^2+4)(s^2 +1)}$

So $\displaystyle A+B=0 \Leftrightarrow A=-B$ and $\displaystyle A+4B=3$ so $\displaystyle A=-1, B=1$ and $\displaystyle C=D=0$

So:

$\displaystyle \frac{3s}{(s^2+4)\cdot(s^2+1}=\frac{-s}{s^2+4}+\frac{s}{s^2+1}$

Check:

$\displaystyle \frac{-s}{s^2+4}+\frac{s}{s^2+1}=\frac{-s(s^2+1)+s(s^2+4)}{(s^2+4)\cdot(s^2+1)}=\frac{-s^3-s+s^3+4s}{(s^2+4)\cdot (s^2+1)}=\frac{3s}{(s^2+4)\cdot(s^2+1)}$

It __seems__ to work.