Calcule f(t) onde:
L^{-1}\left\{\frac{(s+1)^{3}}{s^{4}}\right\}

Passo 1

O enunciado pede calcular
f(t)=\mathcal{L}^{-1}\Bigl\{\frac{(s+1)^{3}}{s^{4}}\Bigr\}.
Para isso, vamos usar a propriedade linear da transformada inversa e o fato de que
\mathcal{L}^{-1}\bigl\{\tfrac{1}{s^{n}}\bigr\}=\frac{t^{\,n-1}}{(n-1)!}.

Passo 2

Expandimos o numerador:
(s+1)^{3}=s^{3}+3s^{2}+3s+1.
Logo
\frac{(s+1)^{3}}{s^{4}} =\frac{s^{3}}{s^{4}}+3\,\frac{s^{2}}{s^{4}}+3\,\frac{s}{s^{4}}+\frac{1}{s^{4}} =\frac{1}{s}+\frac{3}{s^{2}}+\frac{3}{s^{3}}+\frac{1}{s^{4}}.

Passo 3

Agora aplicamos a transformada inversa a cada termo separadamente:
\mathcal{L}^{-1}\!\Bigl\{\tfrac{1}{s}\Bigr\}=1,\quad \mathcal{L}^{-1}\!\Bigl\{\tfrac{3}{s^{2}}\Bigr\}=3\,t,\quad \mathcal{L}^{-1}\!\Bigl\{\tfrac{3}{s^{3}}\Bigr\}=3\,\frac{t^{2}}{2!}=\frac{3}{2}t^{2},\quad \mathcal{L}^{-1}\!\Bigl\{\tfrac{1}{s^{4}}\Bigr\}=\frac{t^{3}}{3!}=\frac{1}{6}t^{3}.

Passo 4

Somando os resultados, obtemos
f(t)=1+3t+\frac{3}{2}t^{2}+\frac{1}{6}t^{3}.

Resposta

f(t)=1+3t+\tfrac{3}{2}t^{2}+\tfrac{1}{6}t^{3}.

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