What is inverse oflaplace 1/((s^2+3s+2)(s+1)) ?

The answer to inverse oflaplace 1/((s^2+3s+2)(s+1)) is -e^{-t}+e^{-t}t+e^{-2t}

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inverse oflaplace 1/((s^2+3s+2)(s+1))

inverse laplace

\frac{1}{( s ^{2} + 3 s + 2 ) ( s + 1 )}

- e^{- t} + e^{- t} t + e^{- 2 t}

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 3 s + 2 ) ( s + 1 )}}

Take the partial fraction of

\frac{1}{( s ^{2} + 3 s + 2 ) ( s + 1 )} : - \frac{1}{s + 1} + \frac{1}{( s + 1 ) ^{2}} + \frac{1}{s + 2}

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

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

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

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

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

= - e^{- t} + e^{- t} t + e^{- 2 t}

y = \frac{x ^{3}}{3} - x^{2} - 3 x

\int \frac{x ^{2} + 4 x - 7}{x - 1} d x

\frac{d}{d x} (\frac{8 x - 1}{6 x + 7})

\int - ln (x) - \frac{ln ( x )}{4} d x

\int_{- 2}^{4} t^{2} - 2 t - 3 d t

What is inverse oflaplace 1/((s^2+3s+2)(s+1)) ?
The answer to inverse oflaplace 1/((s^2+3s+2)(s+1)) is -e^{-t}+e^{-t}t+e^{-2t}