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

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

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

inverse laplace

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

H (t) - e^{- t}

Solution steps

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

Take the partial fraction of

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

\int \frac{x ^{2}}{16 + 9 x ^{2}} d x

\int_{1}^{2} 2 x - x^{2} d x

\int \frac{1}{x ^{2} - 2 x + 37} d x

\frac{d x}{d t} + x^{2} = x

\frac{\partial}{\partial y} (ln (x + 2 y))

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