What is inverse oflaplace 3/((s+1)^2+9) ?

The answer to inverse oflaplace 3/((s+1)^2+9) is e^{-t}sin(3t)

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inverse oflaplace 3/((s+1)^2+9)

inverse laplace

\frac{3}{( s + 1 ) ^{2} + 9}

e^{- t} sin (3 t)

Solution steps

L^{- 1} {\frac{3}{( s + 1 ) ^{2} + 9}}

Apply inverse transform rule:

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

then

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

For

\frac{3}{( s + 1 ) ^{2} + 9} : a = - 1, F (s) = \frac{3}{s ^{2} + 9}

= e^{- t} L^{- 1} {\frac{3}{s ^{2} + 9}}

L^{- 1} {\frac{3}{s ^{2} + 9}} : sin (3 t)

= e^{- t} sin (3 t)

\frac{d}{d x} (tan (13 x))

2 e^{4 x} cos (5 x)

\frac{\partial}{\partial x} (ln (x^{7} - y^{3}))

f (x) = e^{\frac{a}{x}}

What is inverse oflaplace 3/((s+1)^2+9) ?
The answer to inverse oflaplace 3/((s+1)^2+9) is e^{-t}sin(3t)