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

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

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inverse oflaplace 1-1/(1+x)

inverse laplace

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

δ (t) - e^{- t}

Solution steps

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}{1 + x}}

L^{- 1} {1} : δ (t)

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

= δ (t) - e^{- t}

\frac{d}{d x} (4 x^{2} 5 x + 1)

\int \frac{5}{4 sin ( x ) + 3 cos ( x )} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = 0, y (1) = 0, y^{^{'}} (1) = - 4

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

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