Inverse Laplace transformation of (s^2 + s)/(s^2 +1)(s^2 + 2s + 2)

Laplace transform
asked 2021-06-11

Inverse Laplace transformation

\( (s^2 + s)/(s^2 +1)(s^2 + 2s + 2)\)

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\(1.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s-3}\)
\(2.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s-3}\)
\(3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}\)
\(4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}\)