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

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

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

inverse laplace

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

e^{- t} \cdot 2 cos (2 t)

Solution steps

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

Expand

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

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

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

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

= e^{- t} \cdot 2 cos (2 t)

(1 + x)^{- 2}

\int_{0}^{\frac{π}{4}} 3 sec^{2} (x) d x

\int \frac{( x ^{2} - 5 x ) ^{2}}{x} d x

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

f (x) = 2 x + 45, at x = 2

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