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

The answer to inverse oflaplace s/(s+3) is \delta(t)-3e^{-3t}

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inverse oflaplace s/(s+3)

inverse laplace

\frac{s}{s + 3}

δ (t) - 3 e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{s}{s + 3} : 1 - \frac{3}{s + 3}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{3}{s + 3}} : 3 e^{- 3 t}

= δ (t) - 3 e^{- 3 t}

ln (1 + e^{x})

\frac{d}{d x} (arcsin (2 x^{4}))

\int \frac{x ^{2} + 4 x + 7}{x ^{2} - 6 x + 5} d x

\int_{0}^{t} e^{- x} d x

\int (\frac{( x ^{3} + 1 ) ^{2}}{x ^{2}}) d x

What is inverse oflaplace s/(s+3) ?
The answer to inverse oflaplace s/(s+3) is \delta(t)-3e^{-3t}