What is inverse oflaplace (e^{-s}+s)/(s^2+s) ?

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

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inverse oflaplace (e^{-s}+s)/(s^2+s)

inverse laplace

\frac{e ^{- s} + s}{s ^{2} + s}

H (t - 1) (1 - e^{- t + 1}) + Undefined

Solution steps

L^{- 1} {\frac{e ^{- s} + s}{s ^{2} + s}}

Expand

= L^{- 1} {\frac{s}{s ^{2} + s} + \frac{e ^{- s}}{s ^{2} + 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{e ^{- s}}{s ^{2} + s}} + L^{- 1} {\frac{s}{s ^{2} + s}}

L^{- 1} {\frac{e ^{- s}}{s ^{2} + s}} : H (t - 1) (1 - e^{- t + 1})

L^{- 1} {\frac{s}{s ^{2} + s}} :

Undefined

= H (t - 1) (1 - e^{- t + 1}) + Undefined

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

\int_{- a}^{a} \int_{0}^{1} sin (y) e^{x^{2}} d x d y

f (x) = x 1 - x^{2}

\int_{1}^{8} \frac{1}{3 x + 1} d x

x \to - 1 lim (\frac{1}{x + 1})

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