Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$

Question

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!

Answer 1

$$\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {As+B}{s^2+16}+\frac {C}{s-3}+\frac{D}{s+3}$$

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 $$\frac {25s}{(s^2+16)(s-3)(s+3)}=\frac {-s}{s^2+16}+\frac {1}{2(s-3)} + \frac {1}{2(s+3)} $$

Answer 2

You must write $$\frac{25s}{(s^2+16)(s-3)(s+3)}=\frac{As+B}{s^2+16}+\frac{C}{s-3}+\frac{D}{s+3}$$

Answer 3

Your starting point is wrong. When you have a quadratic denominator for one of the partial fractions you need to give it a linear numerator, so you needed to start with $\frac {As+B}{s^2+16}+\frac {C}{s-3}+\frac{D}{s+3}$.