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

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

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

inverse laplace

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

H (t) - 2 e^{- t}

Solution steps

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

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

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

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

Refine

H (t) - e^{- t} - e^{- t} : H (t) - 2 e^{- t}

= H (t) - 2 e^{- t}

\int \frac{1}{x ^{2}} 6 + \frac{1}{x} d x

x \to 0 lim (cot (4 x))

\frac{d}{d x} (\frac{2 x ^{2} - 1}{x ^{3} - x})

\frac{1}{2 r} + \frac{1}{9 r ^{(\frac{8}{9})}}

x \to 0 lim (\frac{1}{3 x ^{3}})

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