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?
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$\begingroup$Recall $e^{a+b}=e^a e^b$. Using linearity of an integral transform you see
$\endgroup$ $\begingroup$$$\mathcal{L}\left\lbrace e^{-2t+1}\right\rbrace = e\int_0^\infty e^{-t(2+s)}\,dt = {e\over 2+s}$$
First of all, $e^{-2t + 1} = e^{-2t}e $
Then, the Laplace transformation :
$\mathcal{L}[e^{-2t + 1}] = \mathcal{L} [e^{-2t}e] = e\mathcal{L} [e^{-2t}] = \frac{e}{s+2}$ by the known Laplace transformation formulas.
Otherwise, just take the integral : $\mathcal{L}\left\lbrace e^{-2t+1}\right\rbrace = e\int_0^\infty e^{-t(2+s)}\,dt = {e\over 2+s}$
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