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

The answer to inverse oflaplace 1/(x(x^2+2x+x)) is -1/9 \H(t)+t/3+1/9 e^{-3t}

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

inverse laplace

\frac{1}{x ( x ^{2} + 2 x + x )}

- \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

Solution steps

L^{- 1} {\frac{1}{x ( x ^{2} + 2 x + x )}}

Take the partial fraction of

\frac{1}{x ( x ^{2} + 2 x + x )} : - \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )}

= L^{- 1} {- \frac{1}{9 x} + \frac{1}{3 x ^{2}} + \frac{1}{9 ( x + 3 )}}

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}{9 x}} + L^{- 1} {\frac{1}{3 x ^{2}}} + L^{- 1} {\frac{1}{9 ( x + 3 )}}

L^{- 1} {\frac{1}{9 x}} : \frac{1}{9} H (t)

L^{- 1} {\frac{1}{3 x ^{2}}} : \frac{t}{3}

L^{- 1} {\frac{1}{9 ( x + 3 )}} : \frac{1}{9} e^{- 3 t}

= - \frac{1}{9} H (t) + \frac{t}{3} + \frac{1}{9} e^{- 3 t}

f (x) = 3 ln (x)

\frac{d}{d x} (\frac{1}{6 x + 5})

\frac{\partial}{\partial y} (7 ln (2 x^{5} - 3 y))

\frac{1}{10 x ^{2}}, x, \frac{x}{10}

\frac{d y}{d x} = k y

What is inverse oflaplace 1/(x(x^2+2x+x)) ?
The answer to inverse oflaplace 1/(x(x^2+2x+x)) is -1/9 \H(t)+t/3+1/9 e^{-3t}