Inverse Laplace Transform of $\frac{s-3}{s[(s-3)^2+9]}$

Question

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 $\frac{1}{s^2-6s+18}$.

Answer 1

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: $$ \mathcal{L}^{-1}\left(\frac{s-3}{s\left((s-3)^2+9\right)}\right)=\color{red}{\frac{1}{6}\left(-1+e^{3t}\left(\sin(3t)+\cos(3 t)\right)\right)}. \tag{4} $$

Answer 2

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 $C=0$.