Inverse Laplace Transform of $\frac{s-3}{s[(s-3)^2+9]}$ I am having problems solving this inverse Laplace transform:
ℒ$^{-1}\Large [\frac{s-3}{s[(s-3)^2+9]}]$
I did partial fraction decomposition, but ended up with complex expressions in some denominators, and I have no idea what do do from here or if this is even how to "solve" it.
Could anyone give me some helpful hints or references? I would appreciate it a lot!
Edit:
My problem is, specifically, finding the inverse Laplace of $\frac{1}{s^2-6s+18}$.
 A: Note $\dfrac{s-3}{s\left[(s-3)^2+9\right]}$ can be expressed as
\begin{align*}
\frac{s-3}{s\left[(s-3)^2+9\right]}&=\frac{s-3}{s(s^2-6s+18)}\\
&=\frac{A}{s}+\frac{Bs+C}{s^2-6s+18}
\end{align*}
For some real numbers $A$, $B$ and $C$, that means
\begin{align*}
s-3&=A(s^2-6s+18)+(Bs+C)s\\
&=(A+B)s^2+(-6A+C)s+18A\\
\end{align*}
$$\iff \left\{\begin{matrix}A+B&=0\\-6A+C&=1\\18A&=-3\end{matrix}\right.$$
Then $A=-\frac{1}{6}$, $B=\frac{1}{6}$ and $C=0$.
A: It is better to exploit some properties of the (inverse) Laplace transform. In particular:
$$\mathcal{L}^{-1}\left(\frac{f(s)}{s}\right) = \int_{0}^{t}\left(\mathcal{L}^{-1} f\right)(u)\,du,\tag{1} $$
$$\mathcal{L}^{-1} g'(s) = -t\cdot \left(\mathcal{L}^{-1} g\right)(t). \tag{2}$$
Since:
$$ \mathcal{L}^{-1}\left(\frac{s-3}{(s-3)^2+9}\right)=e^{3t}\cos(3t), \tag{3} $$
by $(1)$ we simply have:
$$ \mathcal{L}^{-1}\left(\frac{s-3}{s\left((s-3)^2+9\right)}\right)=\color{red}{\frac{1}{6}\left(-1+e^{3t}\left(\sin(3t)+\cos(3 t)\right)\right)}. \tag{4} $$
