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

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

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inverse oflaplace 1/((s+3))

inverse laplace

\frac{1}{( s + 3 )}

e^{- 3 t}

Solution steps

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

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

= e^{- 3 t}

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

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

\frac{d}{d x} (\frac{1}{1 - x})

\int coth^{2} (x) d x

\frac{1}{( x ^{2} + 4 ) ^{3}}

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