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

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

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

inverse laplace

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

H (t) - e^{- t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - e^{- t}

y = x, y = \frac{x}{3}, y = 2

\int_{0}^{2} x^{2} 1 + x^{3} d x

\frac{d}{d x} (\frac{2}{1 + 4 x ^{2}})

\int \frac{x ^{2}}{36 + x ^{2}} d x

x^{3} - 5

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