What is inverse oflaplace (s+1)/(s^2+2s+5) ?

The answer to inverse oflaplace (s+1)/(s^2+2s+5) is e^{-t}cos(2t)

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inverse oflaplace (s+1)/(s^2+2s+5)

inverse laplace

\frac{s + 1}{s ^{2} + 2 s + 5}

e^{- t} cos (2 t)

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} + 2 s + 5}}

\frac{s + 1}{s ^{2} + 2 s + 5} = \frac{s + 1}{( s + 1 ) ^{2} + 4}

= L^{- 1} {\frac{s + 1}{( s + 1 ) ^{2} + 4}}

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} + 4} : a = - 1, F (s) = \frac{s}{s ^{2} + 4}

= e^{- t} L^{- 1} {\frac{s}{s ^{2} + 4}}

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

= e^{- t} cos (2 t)

\frac{d}{d x} (\frac{x ^{2}}{9 - x ^{2}})

\frac{\partial}{\partial x} (2 y x)

\int \frac{x ^{2}}{x ^{2} - x + 1} d x

f (x) = \frac{( x + 1 ) ( 3 x - 7 )}{3 x + 1}

\int x^{4} 9 + x^{5} d x

What is inverse oflaplace (s+1)/(s^2+2s+5) ?
The answer to inverse oflaplace (s+1)/(s^2+2s+5) is e^{-t}cos(2t)