What is inverse oflaplace x/(x-1) ?

The answer to inverse oflaplace x/(x-1) is \delta(t)+e^t

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inverse oflaplace x/(x-1)

inverse laplace

\frac{x}{x - 1}

δ (t) + e^{t}

Solution steps

L^{- 1} {\frac{x}{x - 1}}

Take the partial fraction of

\frac{x}{x - 1} : 1 + \frac{1}{x - 1}

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

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1}{x - 1}} : e^{t}

= δ (t) + e^{t}

\int \frac{x}{e ^{7 x}} d x

x \to 1 lim (\frac{2}{5} x + 7)

\int ln (3)

\frac{\partial}{\partial x} (\frac{6}{x ^{2}})

\int \frac{x ^{3}}{2 - x ^{4}} d x

What is inverse oflaplace x/(x-1) ?
The answer to inverse oflaplace x/(x-1) is \delta(t)+e^t