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

The answer to inverse oflaplace 1/((x^2+1)*x) is -cos(t)+\H(t)

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

inverse laplace

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

- cos (t) + H (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= - cos (t) + H (t)

\int x sin (\frac{1}{4} x) d x

\frac{\partial}{\partial s} (3 s^{2} + t^{2})

\frac{d y}{d x} = x - y, y (0) = 2

x \to 0 lim (ln (\frac{1}{x ^{2}}))

What is inverse oflaplace 1/((x^2+1)*x) ?
The answer to inverse oflaplace 1/((x^2+1)*x) is -cos(t)+\H(t)