What is inverse oflaplace (s+3)/((s+3)^2+25) ?

The answer to inverse oflaplace (s+3)/((s+3)^2+25) is e^{-3t}cos(5t)

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inverse oflaplace (s+3)/((s+3)^2+25)

inverse laplace

\frac{s + 3}{( s + 3 ) ^{2} + 25}

e^{- 3 t} cos (5 t)

Solution steps

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 25}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{s + 3}{( s + 3 ) ^{2} + 25} : a = - 3, F (s) = \frac{s}{s ^{2} + 25}

= e^{- 3 t} L^{- 1} {\frac{s}{s ^{2} + 25}}

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

= e^{- 3 t} cos (5 t)

\int sec^{3} (6 x) d x

- x, x, - 2 x + 6

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

\frac{\partial}{\partial x} (- 5 e^{x} ln (y))

y = x^{2}, y = 3 x + 4

What is inverse oflaplace (s+3)/((s+3)^2+25) ?
The answer to inverse oflaplace (s+3)/((s+3)^2+25) is e^{-3t}cos(5t)