Question

how to find inverse?

Answer 1

\[L^{-1}\begin{Bmatrix}
\frac{2s+6}{(s^2+6s+10)^2}
\end{Bmatrix}\]

Answer 2

\[L^{-1}=laplace\]

Answer 3

I think that it's the laplace \[(-1)^1\frac{ d }{ ds }\frac{ 1 }{ (s^2+6s+10) } \Rightarrow -tL^{-1}\left\{ \frac{ 1 }{ (s+6s+10) } \right\} \Rightarrow -tL \left\{ \frac{1}{(s+3)^2+1} \right\}\]

Answer 4

I think that's a shift theroem where \(s \rightarrow (s+3)\) and which is the inverse laplace of sin(t)
so inverse laplace is: \(-te^{-3t}sin(t)\)