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

The answer to inverse oflaplace ((s^2-1))/((s^2+1)^2) is tcos(t)

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inverse oflaplace ((s^2-1))/((s^2+1)^2)

inverse laplace

\frac{( s ^{2} - 1 )}{( s ^{2} + 1 ) ^{2}}

t cos (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= sin (t) - (sin (t) - t cos (t))

Simplify

sin (t) - (sin (t) - t cos (t)) : t cos (t)

= t cos (t)

x \to 1 - lim ((x^{2} + 1)^{2})

δ (t - 1)

\int_{0}^{π} cos (x + π) d x

\int \frac{x ^{2}}{x} d x

\frac{\partial}{\partial x} (3 a x^{2} + e^{xx} + e^{x})

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