Inverse laplace transform $1/(s^2+9)^2$ 
Find inverse laplace transform of $$\frac{1}{(s^2+9)^2}$$

I've tried to decompose the fraction using
$$\frac{As+B}{s^2+9}+\frac{Cs+D}{(s^2+9)^2}$$
$$1=(As+B)(s^2+9)+Cs+D$$
yet D=1, still giving me the same exact equation
$$\frac{1}{(s^2+9)^2}$$
any help?
 A: You can pull this straight from the table
$$\mathcal{L}^{-1} \left(\frac{1}{(s^2 + k^2)^2}\right) = \frac{1}{2k^3}(\sin(kt) - kt\cos(kt)).$$
The answer is
$$\frac{1}{54}(\sin{3t} - 3t\cos{3t}).$$
A: Rewrite it as follows
$$\frac{1}{(s^2+9)^2}=-\frac{1}{2s}\frac{-2s}{(s^2+9)^2}=-\frac{1}{2s}\frac{\mathrm d}{\mathrm ds}\left( \frac{1}{s^2+9}\right)$$
Now use the following properties:

$$tf(t)\stackrel{\mathcal{L}}\longleftrightarrow-\frac{\mathrm d}{\mathrm ds}F(s)$$
  $$\int_{0}^{t}g(\tau)d\tau\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s}G(s)$$

as well as 

$$\frac{1}{3}\sin(3t)\stackrel{\mathcal{L}}\longleftrightarrow\frac{1}{s^2+9}$$

A: One could make an ansatz with "known" expressions from a simple Laplace Transform Table:
$$\frac1{(s^2+a^2)^2}=\frac{Aa}{s^2+a^2}+\frac{Bs}{s^2+a^2}+\frac{2Cs}{(s^2+a^2)^2}+\frac{D(s^2-a^2)}{(s^2+a^2)^2},$$
given the Laplace transforms of $\sin at$, $\cos at$, $t\sin at$, and $t\cos at$, respectively. Hence, we get $A=1/(2a^3)$, $B=C=0$, and $D=-1/(2a^2)$. Therefore
$$\frac1{(s^2+a^2)^2}=\frac1{2a^3}\cdot\frac a{s^2+a^2}-\frac1{2a^2}\cdot\frac{s^2-a^2}{(s^2+a^2)^2}.$$
