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

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

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

inverse laplace

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

e^{- 2 t} sin (t)

Solution steps

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

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

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

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

\frac{8 - x e ^{x}}{x + e ^{x}}

\int \frac{x + x ^{3}}{x} - e^{x} d x

f (x) = \frac{e ^{x} - e ^{- x}}{2}

\int x^{3} e^{3 x} d x

\int 2 t + 4 d t

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