TITLE: Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$
QUESTION [2 upvotes]: So this is the problem..
$$
\frac{25s}{(s^2+16)(s-3)(s+3)}
$$
So what I did was...
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {A}{s^2+16}+\frac {B}{s-3}+\frac{C}{s+3}
$$
then...
$$\begin{align}
25s &= A(s-3)(s+3)+B(s^2+16)(s+3)+C(s^2+16)(s-3) \\
25s &= A(s^2-9)+B(s^3+3s^2+16s+48)+C(s^3-3s^2+16s-48) \\
25s &= As^2-9A+Bs^3+3Bs^+16Bs+48B+Cs^3-3Cs^2+16Cs-48C
\end{align}$$  
rearranged to get... 
$$
25s = Bs^3+Cs^3+As^2+3Bs^2-3Cs^2+16Bs+16Cs-9A+48B-48C
$$   
factoring out...  
$$
25s = (B+C)s^3+(A+3B-3C)s^2+(16B+16C)s-9A+48B-48C
$$   
Now, this is the part where U messed up. I've been trapped with this question for more than an hour. I don't know if I made any mistakes as I've looked at it over and over again.
The answer is supposed to be:
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {-s}{s^2+16}+\frac {1}{2(s-3)}+\frac{1}{2(s+3)}
$$  
Please be detailed. Thank you!

REPLY [3 votes]: $$\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {As+B}{s^2+16}+\frac {C}{s-3}+\frac{D}{s+3}$$
Use Heaviside method.
To find $C$, let $s=3$ and you get  $C=1/2$ 
To find $D$, let $s=-3$ and you get  $D=1/2$ 
Substitute for $C$ and $D$ in the RHS and subtract from LHS.
You will find $$\frac {25s}{(s^2+16)(s-3)(s+3)}-\frac {1}{2(s-3)} - \frac {1}{2(s+3)}= \frac {-s}{s^2+16}$$
Thus $$\frac {25s}{(s^2+16)(s-3)(s+3)}=\frac {-s}{s^2+16}+\frac {1}{2(s-3)} + \frac {1}{2(s+3)} $$
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