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

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

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

inverse laplace

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

e^{- t} cos (t)

Solution steps

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

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

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

Apply inverse transform rule:

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

then

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

For

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

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

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

= e^{- t} cos (t)

cos^{3} (t)

\frac{d y}{d x} = \frac{6 y e ^{3 x}}{2 + e ^{3 x}}

\int_{- \infty}^{- 1} e^{- 28 t} d t

\frac{d}{d x} (x (x - 1)^{2} (x - 2)^{3})

\int π^{π} d x

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