F(s) = s+1/(s^2 + 2s + 10)^2 Find the La place inverse transform
You have $\displaystyle F(s) = \frac{s + 1}{([s + 1]^2 + 3^2)^2} = G(s + 1)$ where $\displaystyle G(s) = \frac{s}{(s^2 + 3^2)^2}$.
From a well known theorem: $\displaystyle LT^{-1}[G(s + 1) = e^t g(t)$ where $\displaystyle g(t) = LT^{-1}[G(s)]$.
To get g(t), first note that $\displaystyle G(s) = \frac{dK}{ds}$ where $\displaystyle K(s) = - \frac{1}{2} \left(\frac{1}{s^2 + 3^2} \right)$.
Now use the well known theorem that $\displaystyle LT^{-1}\left[\frac{dK}{ds}\right] = - t k(t)$ where $\displaystyle k(t) = LT^{-1}[K(s)]$.