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The point is that all three methods:
a)[tex]\displaystyle \mathcal{L}[e^{at}f(t)] = F(s-a),[/tex]
where F(s) is the Laplace transform of t.
b)[tex]\displaystyle \mathcal{L}[tf(t)]= -F'(s)[/tex]
where F(s) is the Laplace transform of [itex]e^{3t}[/itex].
c)[tex]\displaystyle \mathcal{L}[te^{at}]= \int_0^\infty te^{(-s+3)t}dt[/tex]
will give the same result.
It would be a good exercise to try each method and see.
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