What is the inverse Laplace transform of squared denominator term? Referring to the image below, what would the inverse Laplace transform be? I can't seem to find any tables that include this case.
This is quite straightforward to solve. Either just use this table where you can directly find the corresponding result, or "derive" it yourself with very basic knowledge of the Laplace transform.

You should know that

$$\mathcal{L}\{u(t)\}=\frac{1}{s}\tag{1}$$

Integration is equivalent to multiplication with $1/s$, so integrating $(1)$ gives

$$\mathcal{L}\{t\cdot u(t)\}=\frac{1}{s^2}\tag{2}$$

Replacing $s$ by $s+a$ is equivalent to multiplication with $e^{-at}$. Consequently, we have

$$\mathcal{L}^{-1}\left\{\frac{2}{(s+2)^2}\right\}=2\,t\,e^{-2t} u(t)\tag{3}$$