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)

h \to 0 lim (h - \frac{9}{h} + 4)

\int \frac{x ^{2} - 5 x + 2}{3 x} d x

f (x) = - 2 x^{3} + x^{2}, at x = 1

y = 10 x, (10, 10)

\frac{d}{d x} (ln (5 - x^{2}))

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)