What is inverse oflaplace ((s+1)^2)/s ?

The answer to inverse oflaplace ((s+1)^2)/s is \delta'(t)+2\delta(t)+\H(t)

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inverse oflaplace ((s+1)^2)/s

inverse laplace

\frac{( s + 1 ) ^{2}}{s}

δ' (t) + 2 δ (t) + H (t)

Solution steps

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

Take the partial fraction of

\frac{( s + 1 ) ^{2}}{s} : s + 2 + \frac{1}{s}

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

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {2} : 2 δ (t)

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

= δ^{^{'}} (t) + 2 δ (t) + H (t)

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

\frac{d}{d x} ((x - 8)^{3})

x \to 0 + lim ([x - sin (x)])

x \to 1 + lim ((x - 1)^{l n (x)})

\frac{1}{x + 9}

What is inverse oflaplace ((s+1)^2)/s ?
The answer to inverse oflaplace ((s+1)^2)/s is \delta'(t)+2\delta(t)+\H(t)