I need to do the inverse Laplace transform of such fraction: $$\frac{\frac{3}{9 + (-1 + p)^2}+p-3}{p^{2} -2p+10}$$. I've thought of rewriting it as $$\frac{\frac{3}{9 + (-1 + p)^2}+p-3}{p^{2} -2p+10}=\frac{\frac{3}{p^2 - 2p + 10}+p-3}{p^{2} -2p+10}=\frac{3}{(p^{2} -2p+10)^2}+\frac{p}{p^{2} -2p+10}-\frac{3}{p^{2} -2p+10}$$, but I don't know which original functions can I get from those fractions and stuck with that. Maybe there is a better way to expand it to particle fractions?
Tried using Wolfram Alpha and Symbolab to figure a solution for this - neither of them can show step by step solution.
$$\cfrac{\frac{3}{9+(-1+p)^2}+p+3}{p^2-2p+10}=\underbrace{\frac{3}{(p^2-2p+10)^2}}_{\color{red}{1}}+\underbrace{\frac{p+3}{p^2-2p+10}}_{\color{red}{2}}$$
Lets look at $\color{red}{2}$ first: $$\frac{p+3}{p^2-2p+10}=\frac{(p-1)+4}{(p-1)^2+3^2}=\frac{(p-1)}{(p-1)^2+3^2}+\frac{4}{(p-1)^2+3^2}$$ Now you will want to make use of the fact that: $$\mathscr{L}\left[e^{at}f(t)\right]\{s\}=\mathscr{L}[f(t)]\{s-a\}$$ or you may prefer to write it as: $$\mathscr{L}[f(t)]=F(s),\,\,\mathscr{L}\left[e^{at}f(t)\right]=F(s-a)$$ which tells us that: $$\mathscr{L}^{-1}\left[\frac{p-1}{(p-1)^2+3^2}\right]=e^t\mathscr{L}^{-1}\left[\frac{p}{p^2+3^2}\right]=e^t\cos 3t$$
Now for the first one, using what we know before: $$\mathscr{L}^{-1}\left[\frac{3}{((p-1)^2+3^2)^2}\right]=e^t\mathscr{L}^{-1}\left[\frac{3}{(p^2+3^2)^2}\right]$$ so you just need to find the transform of this now.
Hint: $$\mathscr{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at$$