What is inverse oflaplace (x+1)/(x^2-4x) ?

The answer to inverse oflaplace (x+1)/(x^2-4x) is -1/4 \H(t)+5/4 e^{4t}

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inverse oflaplace (x+1)/(x^2-4x)

inverse laplace

\frac{x + 1}{x ^{2} - 4 x}

- \frac{1}{4} H (t) + \frac{5}{4} e^{4 t}

Solution steps

L^{- 1} {\frac{x + 1}{x ^{2} - 4 x}}

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{4 x} + \frac{5}{4 ( x - 4 )}}

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}{4 x}} + L^{- 1} {\frac{5}{4 ( x - 4 )}}

L^{- 1} {\frac{1}{4 x}} : \frac{1}{4} H (t)

L^{- 1} {\frac{5}{4 ( x - 4 )}} : \frac{5}{4} e^{4 t}

= - \frac{1}{4} H (t) + \frac{5}{4} e^{4 t}

\int csc^{2} (x) + 7 d x

x \to 6 lim (x^{2} - 7 x + 6)

\frac{\partial}{\partial x} (5 y ln (x))

\frac{d}{d x} (\frac{x ^{3}}{3 - x ^{2}})

\frac{1}{2 x + 10}

What is inverse oflaplace (x+1)/(x^2-4x) ?
The answer to inverse oflaplace (x+1)/(x^2-4x) is -1/4 \H(t)+5/4 e^{4t}