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

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

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inverse oflaplace 1/(1+s)

inverse laplace

\frac{1}{1 + s}

e^{- t}

Solution steps

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

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

= e^{- t}

\frac{d}{d x} (\frac{arcsin ( x )}{x})

\frac{\partial}{\partial x} (x^{2} y + 4 y)

\int \frac{- 5 x}{( x ^{2} + 5 ) ^{\frac{3}{2}}} d x

y = x ln (sin^{2} (x))

y (x + 7) + \frac{d y}{d x} = 0

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