What is inverse oflaplace 1/(s^2+2s+3) ?

The answer to inverse oflaplace 1/(s^2+2s+3) is e^{-t} 1/(sqrt(2))sin(sqrt(2)t)

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inverse oflaplace 1/(s^2+2s+3)

inverse laplace

\frac{1}{s ^{2} + 2 s + 3}

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

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 2 s + 3}}

\frac{1}{s ^{2} + 2 s + 3} = \frac{1}{( s + 1 ) ^{2} + 2}

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

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

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

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

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

\int_{2}^{9} \frac{1}{7 ( x - 2 ) ^{3}} d x

x \to - 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

\frac{\partial}{\partial x} (\frac{x}{( x - 1 ) ^{2} + y ^{2}})

\int_{5}^{53} 100 - x^{2} d x

\int e^{\frac{1}{2}}

What is inverse oflaplace 1/(s^2+2s+3) ?
The answer to inverse oflaplace 1/(s^2+2s+3) is e^{-t} 1/(sqrt(2))sin(sqrt(2)t)