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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int \frac{1}{x - 6} d x

- 1 {\frac{- 3}{- s + 2}}

\int \frac{z}{2 z ^{2} + 8} d z

(- 2, 3), (- 1, - 2)

\frac{d}{d x} ((\frac{a x + b}{c ^{2} + a})^{10})

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