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

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

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

inverse laplace

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

e^{- t} cos (2 t)

Solution steps

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

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

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

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

\int x^{2} 4 x + 8 d x

y = x^{2}, y = 4 + x

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

y = x^{2}, y = x + 2, x = - 2

ln (\frac{1}{x + 5}) + x cos (x)

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