I've been trying to prove the result $$\mathcal{L}^{-1} \{\frac{3}{(s^2+9)^2}\}(t)=\frac{1}{18} (\sin(3t)-3t\cos(3t))$$
by a contour integral. So $$\begin{align}\mathcal{L}^{-1} \{\frac{3}{(s^2+9)^2}\}(t) = \end{align}\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}{\frac{3e^{st}}{(s^2+9)^2}}\, ds $$
(and with poles at $\pm3i$, using a Bromwich contour)
$$\mathcal{L}^{-1} \{\frac{3}{(s^2+9)^2}\}(t)=\sum{\mathrm{Res}{\left(\frac{3e^{st}}{(s^2+9)^2}\right)}}$$
So with $(s^2+9)^2=(s+3i)^2(s-3i)^2$
$$\begin{align} &=3\lim_{s\to3i}\frac{d}{ds}{\left((s-3i)^2 \frac{e^{st}}{(s+3i)^2(s-3i)^2}\right)}+3\lim_{s\to-3i}\frac{d}{ds}{\left((s+3i)^2 \frac{e^{st}}{(s+3i)^2(s-3i)^2}\right)} \\&=3\lim_{s\to3i}{\left(\frac{te^{st}(s+3i)^2-2e^{st}(s+3i)}{(s+3i)^4}\right)}+3\lim_{s\to-3i}{\left(\frac{te^{st}(s-3i)^2-2e^{st}(s-3i)}{(s-3i)^4}\right)} \\&=3\lim_{s\to3i}{\left(\frac{e^{st}(t(s+3i)-2)}{(s+3i)^3}\right)}+3\lim_{s\to-3i}{\left(\frac{e^{st}(t(s-3i)-2)}{(s-3i)^3}\right)} \\&=3\left(\frac{e^{3it}(6it-2)}{-216i}\right)+3\left(\frac{e^{-3it}(-6it-2)}{216i}\right) \\&=3\left(\frac{-6te^{3it}-2ie^{3it}}{216}\right)-3\left(\frac{6te^{-3it}-2ie^{-3it}}{216}\right) \\&=3\left(\frac{-6t(e^{3it}+e^{-3it})-2i(e^{3it}-e^{-3it})}{216}\right) \\&=3\left(\frac{-12t\cos(3t)+4\sin(3t)}{216}\right) \\&=3\left(\frac{1}{54}(\sin(3t))-\frac{1}{18}(t\cos(3t))\right) \\&=\frac{1}{18}\left(\sin(3t)-3t\cos(3t)\right) \end{align}$$
EDIT : Accounted for the 3 in the numerator (and corrected the final line). Proof is all good now.
Forgot about the 3 in the numerator (thanks @J.G.) and also in the final line I lost a $t$ coefficient on cosine.