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

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

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

inverse laplace

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

δ' (t) + 2 δ (t) + 2 H (t)

Solution steps

L^{- 1} ⎩ ⎨ ⎧ \frac{( ( s + 1 ) ^{2} + 1 )}{s} ⎭ ⎬ ⎫

Take the partial fraction of

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

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

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {2} : 2 δ (t)

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

= δ^{^{'}} (t) + 2 δ (t) + 2 H (t)

\int 2 x (x^{2} + 3)^{7} d x

x - e^{3 x}

\int \frac{3}{8 + 6 x} d x

y = \frac{1}{x}, y = \frac{1}{4} x

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