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 ( 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}

f (x) = \frac{1}{2 x ^{2}}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + y ^{2}})

\int_{0}^{1} (3 x - 1)^{50} d x

\int 7 x^{9} e^{- x^{5}} d x

\frac{\partial}{\partial x} (4 e^{x} ln (1 + y))

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