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

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

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

inverse laplace

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

Undefined

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 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{s}{s ^{2} + 2 s} : F (s) = \frac{1}{s ^{2} + 2 s}

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

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

Evaluate

\frac{1}{2} H (t) - \frac{1}{2} e^{- 2 t}

at point

0 =

Undefined

= Undefined

x \to \infty lim (5 + \frac{3}{x})

\int 3 3 - 4 x^{2} (- 8 x) d x

\int 15 tan^{2} (x se) c^{3} x

\frac{d}{d x} (x sin (\frac{x}{2}))

n = 0 \sum \infty n \cdot sin (\frac{1}{n})

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