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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

\int \frac{3 x - 4}{x ^{2} - x - 2} d x

\int \frac{5 x ^{2} + 9 x + 5}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d a} (sin (a) - sin (2 a) + sin (3 a))

x \to 0 - lim (x cot (x))

y = \frac{7}{x ^{3}}

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