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

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

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inverse oflaplace 1/s*1/(s+1)

inverse laplace

\frac{1}{s} \cdot \frac{1}{s + 1}

H (t) - e^{- t}

Solution steps

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

Expand

\frac{1}{s} \frac{1}{s + 1} : \frac{1}{s ^{2} + s}

= L^{- 1} {\frac{1}{s ^{2} + s}}

Take the partial fraction of

\frac{1}{s ^{2} + s} : \frac{1}{s} - \frac{1}{s + 1}

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

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

L^{- 1} {\frac{1}{s}} : H (t)

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

= H (t) - e^{- t}

\int 8 sin^{2} (x + \frac{π}{12}) d x

n = 1 \sum \infty \infty

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

f (x) = \frac{x - 1}{x ^{2}}

\frac{\partial}{\partial x} (3 x^{2} y + 2 x y^{3})

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