Find $\mathcal{L}^{-1}\frac{s}{s^2-6s+9}$

Question

It is easy to see that $\frac{s}{s^2-6s+9}=\frac{s}{(s-3)^2}$ and now I want to use use the convolution integral for $s\cdot \frac{1}{(s-3)^2}$. So I get this integral: $$\int_0^t \delta '(\tau)\cdot \tau e^{3\tau}d\tau$$ Which I din't know how to compute. Is this the correct reasoning at all?

Answer 1

See this table for example:

$$\mathcal{L}^{-1}\frac{s}{s^2-6s+9}=e^{3t}\mathcal{L}^{-1}\frac{s+3}{s^2} =e^{3t}\mathcal{L}^{-1}\left(\frac1s+\frac3{s^2}\right)=(1+3t)e^{3t}$$