What is inverse oflaplace 3/(2(s+1)^2+32) ?

The answer to inverse oflaplace 3/(2(s+1)^2+32) is e^{-t} 3/8 sin(4t)

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inverse oflaplace 3/(2(s+1)^2+32)

inverse laplace

\frac{3}{2 ( s + 1 ) ^{2} + 32}

e^{- t} \frac{3}{8} sin (4 t)

Solution steps

L^{- 1} {\frac{3}{2 ( s + 1 ) ^{2} + 32}}

Apply inverse transform rule:

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

then

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

For

\frac{3}{2 ( s + 1 ) ^{2} + 32} : a = - 1, F (s) = \frac{3}{2 s ^{2} + 32}

= e^{- t} L^{- 1} {\frac{3}{2 s ^{2} + 32}}

L^{- 1} {\frac{3}{2 s ^{2} + 32}} : \frac{3}{8} sin (4 t)

= e^{- t} \frac{3}{8} sin (4 t)

\int tan^{4} (6 x) d x

5 x + 3

\int \frac{10}{x - 5} d x

f (x) = 2 x^{2} - 7 x + 5, at x = 3

\int_{0}^{t} cos (x^{2}) d x

What is inverse oflaplace 3/(2(s+1)^2+32) ?
The answer to inverse oflaplace 3/(2(s+1)^2+32) is e^{-t} 3/8 sin(4t)