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

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

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

inverse laplace

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

δ (t) - 2 e^{- t} cos (t) + e^{- t} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

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

L^{- 1} {1} : δ (t)

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

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

= δ (t) - 2 e^{- t} cos (t) + 1 \cdot e^{- t} sin (t)

Refine

δ (t) - 2 e^{- t} cos (t) + 1 e^{- t} sin (t) : δ (t) - 2 e^{- t} cos (t) + e^{- t} sin (t)

= δ (t) - 2 e^{- t} cos (t) + e^{- t} sin (t)

\int \frac{( e ^{x} )}{x} d x

f (x) 2 x^{2} + 8 x, x = 1

x \to 0 lim ((tan (6 x))^{x})

\frac{\partial}{\partial x} (x + y + x^{3} + y^{2})

y^{^{'}} = x^{2} (x - 2)^{2} (x - 1)^{2}

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