I have the fraction: $$\frac{s}{s^2+2s+2}$$
I want to rearrange the fraction so that I can solve find the inverse Laplace of it using the following identities from a Laplace Transform table:
$$f(t).................F(s)$$ $$d(t)....................1$$ $$tut(t)..................\frac{1}{s}$$ $$t^nu(t)..............\frac{n!}{s^1+1}$$ $$e^{-at}u(t).............\frac{1}{s+a}$$ $$\sin \omega tu(t)............\frac{\omega}{s^2+\omega^2}$$ $$\cos\omega tu(t).............\frac{s}{s^2+\omega^2}$$
I have tried to pull the two out from the bottom however it only makes it more complex and doesn't get me closer to one of the identities. I feel like this should be easy and I am forgetting some high school maths somewhere.
One approach is to note that the denominator is $s^2+2s+2=(s+1)^2+1$, so to define $t=s+1$ and then $$\frac s{s^2+2s+2}=\frac s{(s+1)^2+1}=\frac {t-1}{t^2+1}=\frac t{t^2+1}-\frac 1{t^2+1}$$ and the last two are in your table. If you don't mind complex variables, you can do partial fractions $$\frac s{s^2+2s+2}=\frac s{(s+1)^2+1}=\frac s{(s+1-i)(s+1+i)}$$ and the result will be in the table.