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)

f (x) = x^{2} + 1, at x = - 5

\int_{0}^{1} cos (π x) d x

\frac{d ^{4}}{d x ^{4}} (sin (x))

x \to 3 lim (2 x^{3} - 4 x - 2)

\int x^{35} sin (x^{36}) d x

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)