Laplace transform $e^1$ Sorry if this question seems a little easy to solve, just need a little guidance.
So I have to find the Laplace transform of $e^{-2t+1}$. So I separated the exponential to $e^{-2t} + e^1$. Is the Laplace transform of $e^1 = e/s$ since it is just a constant?

Recall $e^{a+b}=e^a e^b$. Using linearity of an integral transform you see

$$\mathcal{L}\left\lbrace e^{-2t+1}\right\rbrace = e\int_0^\infty e^{-t(2+s)}\,dt = {e\over 2+s}$$

