What is inverse oflaplace 1/((s^2+2s+2)(s+1)) ?

The answer to inverse oflaplace 1/((s^2+2s+2)(s+1)) is -e^{-t}cos(t)+e^{-t}

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inverse oflaplace 1/((s^2+2s+2)(s+1))

inverse laplace

\frac{1}{( s ^{2} + 2 s + 2 ) ( s + 1 )}

- e^{- t} cos (t) + e^{- t}

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 2 s + 2 ) ( s + 1 )}}

Take the partial fraction of

\frac{1}{( s ^{2} + 2 s + 2 ) ( s + 1 )} : \frac{- s - 1}{s ^{2} + 2 s + 2} + \frac{1}{s + 1}

= L^{- 1} {\frac{- s - 1}{s ^{2} + 2 s + 2} + \frac{1}{s + 1}}

Expand

\frac{- s - 1}{s ^{2} + 2 s + 2} : \frac{- ( s + 1 )}{( s + 1 ) ^{2} + 1}

= L^{- 1} {\frac{- ( s + 1 )}{( s + 1 ) ^{2} + 1} + \frac{1}{s + 1}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{- ( s + 1 )}{( s + 1 ) ^{2} + 1}} + L^{- 1} {\frac{1}{s + 1}}

L^{- 1} {\frac{- ( s + 1 )}{( s + 1 ) ^{2} + 1}} : - e^{- t} cos (t)

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

= - e^{- t} cos (t) + e^{- t}

\int 5 sin (6 π t) d t

\int - \frac{1}{2 x + 1} d x

\frac{d}{d x} (\frac{2}{2 x - 1} - \frac{1}{x})

\int \frac{5 x - 3}{x ( x - 3 ) ( x + 1 )} d x

\int x cos (7 π x^{2}) d x

What is inverse oflaplace 1/((s^2+2s+2)(s+1)) ?
The answer to inverse oflaplace 1/((s^2+2s+2)(s+1)) is -e^{-t}cos(t)+e^{-t}