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

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

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inverse oflaplace 3/(s+1)

inverse laplace

\frac{3}{s + 1}

3 e^{- t}

Solution steps

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

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

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

= 3 e^{- t}

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

x \to 0 - lim (e^{\frac{3}{x}})

\frac{\partial}{\partial x} (9)

\frac{d}{d x} ([x + (x + sin^{2} (x))^{3}]^{5})

\int_{0}^{1} 2^{- x} d x

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