Calcule mathscr{L}{f(t)}.
f(t)=\left(1+e^{2 t}\right)^{2}

Passo 1

O enunciado pede a transformada de Laplace de f(t)=(1+e^{2t})^{2}. Vamos usar linearidade da transformada de Laplace e as fórmulas básicas \mathscr{L}\{1\}(s)=\tfrac{1}{s} e \mathscr{L}\{e^{at}\}(s)=\tfrac{1}{s-a}.

Passo 2

Expandindo o quadrado, obtemos
f(t)=(1+e^{2t})^{2}=1+2e^{2t}+e^{4t}.

Passo 3

Aplicando a transformada de Laplace em cada termo e somando:
\mathscr{L}\{1\}(s)=\frac{1}{s},\quad \mathscr{L}\{e^{2t}\}(s)=\frac{1}{s-2},\quad \mathscr{L}\{e^{4t}\}(s)=\frac{1}{s-4}.
Pela linearidade,
\mathscr{L}\{f(t)\}(s) =\frac{1}{s}+2\cdot\frac{1}{s-2}+\frac{1}{s-4}.

Resposta

\mathscr{L}\{\, (1+e^{2t})^{2}\,\}(s) =\frac{1}{s}+\frac{2}{s-2}+\frac{1}{s-4}, \quad \Re(s)>4.

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