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

The answer to inverse oflaplace 1/(x^2(x^2+1)) is t-sin(t)

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

inverse laplace

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

t - sin (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

= t - sin (t)

\frac{\partial}{\partial x} (θ)

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

\frac{37}{x ^{3}}, 1, 100

y^{^{'}} + 8 y = 1, y (0) = 1

f (t) = e^{2 t + 1}

What is inverse oflaplace 1/(x^2(x^2+1)) ?
The answer to inverse oflaplace 1/(x^2(x^2+1)) is t-sin(t)