Calcule as transformadas inversas das seguintes funções:
F(s)=\frac{5 s-2}{3 s^{2}+4 s+8}

Passo 1

O enunciado pede a transformada inversa de Laplace de
F(s)=\frac{5s-2}{3s^{2}+4s+8}.
O conteúdo é “transformada inversa de Laplace” e vamos usar a técnica de completar o quadrado no denominador e depois decompô-lo em termos cujas inversas conhecemos na tabela.

Passo 2

Vamos completar o quadrado em 3s^{2}+4s+8. Primeiro, fatoramos o 3:
3s^{2}+4s+8 = 3\Bigl(s^{2}+\tfrac{4}{3}s+\tfrac{8}{3}\Bigr).
Dentro do parêntese, escrevemos
s^{2}+\tfrac{4}{3}s = \bigl(s+\tfrac{2}{3}\bigr)^{2}-\bigl(\tfrac{2}{3}\bigr)^{2},
logo
s^{2}+\tfrac{4}{3}s+\tfrac{8}{3} = \bigl(s+\tfrac{2}{3}\bigr)^{2}-\tfrac{4}{9}+\tfrac{8}{3} = \bigl(s+\tfrac{2}{3}\bigr)^{2}+\tfrac{20}{9}.
Portanto
3s^{2}+4s+8 = 3\Bigl[(s+\tfrac{2}{3})^{2}+\tfrac{20}{9}\Bigr] =3\Bigl[(s+\tfrac{2}{3})^{2}+\bigl(\tfrac{2\sqrt{5}}{3}\bigr)^{2}\Bigr].

Passo 3

Reescrevemos o numerador 5s-2 em função de (s+\tfrac{2}{3}). Note que
5s-2 = 5\Bigl(s+\tfrac{2}{3}\Bigr) - \tfrac{16}{3}.
Substituindo em F(s):
F(s) = \frac{5(s+\tfrac{2}{3}) - \tfrac{16}{3}} {3\bigl[(s+\tfrac{2}{3})^{2}+(\tfrac{2\sqrt{5}}{3})^{2}\bigr]} = \frac{5}{3}\,\frac{s+\tfrac{2}{3}} {(s+\tfrac{2}{3})^{2}+(\tfrac{2\sqrt{5}}{3})^{2}} \;-\;\frac{16/3}{3\bigl[(s+\tfrac{2}{3})^{2}+(\tfrac{2\sqrt{5}}{3})^{2}\bigr]}.
Simplificando o segundo coeficiente:
\tfrac{16/3}{3}=\tfrac{16}{9}.

Passo 4

Agora usamos as fórmulas conhecidas:
  1. \displaystyle \mathcal{L}^{-1}\!\bigl[\frac{s+a}{(s+a)^{2}+b^{2}}\bigr]=e^{-a t}\cos(bt).
  2. \displaystyle \mathcal{L}^{-1}\!\bigl[\frac{1}{(s+a)^{2}+b^{2}}\bigr]=\frac{1}{b}\,e^{-a t}\sin(bt).
Aqui a=\tfrac{2}{3} e b=\tfrac{2\sqrt{5}}{3}. Então
\mathcal{L}^{-1}\!\Bigl[\frac{s+\tfrac{2}{3}}{(s+\tfrac{2}{3})^{2}+(\tfrac{2\sqrt{5}}{3})^{2}}\Bigr] = e^{-\tfrac{2}{3}t}\cos\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr),
\mathcal{L}^{-1}\!\Bigl[\frac{1}{(s+\tfrac{2}{3})^{2}+(\tfrac{2\sqrt{5}}{3})^{2}}\Bigr] = \frac{1}{\tfrac{2\sqrt{5}}{3}}\,e^{-\tfrac{2}{3}t}\sin\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr) = \frac{3}{2\sqrt{5}}\,e^{-\tfrac{2}{3}t}\sin\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr).

Passo 5

Combinando tudo:
f(t) = \mathcal{L}^{-1}[F(s)] = \frac{5}{3}\,e^{-\tfrac{2}{3}t}\cos\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr) \;-\;\frac{16}{9}\,\cdot\frac{3}{2\sqrt{5}}\,e^{-\tfrac{2}{3}t}\sin\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr).
Simplificando o segundo coeficiente: \tfrac{16}{9}\cdot\tfrac{3}{2\sqrt{5}}=\tfrac{8}{3\sqrt{5}}=\tfrac{8\sqrt{5}}{15}.

Resposta

f(t) = \frac{5}{3}\,e^{-\tfrac{2}{3}t}\cos\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr) \;-\;\frac{8\sqrt{5}}{15}\,e^{-\tfrac{2}{3}t}\sin\!\bigl(\tfrac{2\sqrt{5}}{3}t\bigr).

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