What is inverse oflaplace 1/((s+1)^2+4) ?

The answer to inverse oflaplace 1/((s+1)^2+4) is e^{-t} 1/2 sin(2t)

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inverse oflaplace 1/((s+1)^2+4)

inverse laplace

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

e^{- t} \frac{1}{2} sin (2 t)

Solution steps

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

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

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

= e^{- t} \frac{1}{2} sin (2 t)

\int \frac{x}{x + 3} d x

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What is inverse oflaplace 1/((s+1)^2+4) ?
The answer to inverse oflaplace 1/((s+1)^2+4) is e^{-t} 1/2 sin(2t)