What is inverse oflaplace (2s)/(s^2+s) ?

The answer to inverse oflaplace (2s)/(s^2+s) is Undefined

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inverse oflaplace (2s)/(s^2+s)

inverse laplace

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

Undefined

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{2 s}{s ^{2} + s} : F (s) = \frac{2}{s ^{2} + s}

= \frac{d}{d t} (L^{- 1} {\frac{2}{s ^{2} + s}}) + L^{- 1} {\frac{2}{s ^{2} + s}} (0)

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

Evaluate

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

at point

0 =

Undefined

= Undefined

What is inverse oflaplace (2s)/(s^2+s) ?
The answer to inverse oflaplace (2s)/(s^2+s) is Undefined