What is inverse oflaplace 1/s+1/s e^s ?

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

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inverse oflaplace 1/s+1/s e^s

inverse laplace

\frac{1}{s} + \frac{1}{s} e^{s}

H (t) + H (t + 1)

Solution steps

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

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} e^{s}}

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

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

= H (t) + H (t + 1)

- 2

\int 4 sec (x) tan (x) - 2 sec^{2} (x) d x

\frac{1}{s ^{2} + 3}

\int \frac{x - 1}{x ^{3} - x ^{2} - 2 x} d x

\int 5 sin^{3} (x co) s^{5} x d x

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