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

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

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

inverse laplace

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

e^{- t} cos (t)

Solution steps

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)

(- 1, 6), (7, - 2)

\int \frac{1}{10 x + 1} d x

\frac{d y}{d x} = y - x

F (C) = \frac{9}{5} C + 32

(2 x - 3 y) d x + (2 y - 3 x) d y = 0

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