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

The answer to inverse oflaplace (s^2)/((s^2-1)^2) is -1/4 e^{-t}+(e^{-t}t)/4+1/4 e^t+(e^tt)/4

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

inverse laplace

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

- \frac{1}{4} e^{- t} + \frac{e ^{- t} t}{4} + \frac{1}{4} e^{t} + \frac{e ^{t} t}{4}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {- \frac{1}{4 ( s + 1 )} + \frac{1}{4 ( s + 1 ) ^{2}} + \frac{1}{4 ( s - 1 )} + \frac{1}{4 ( s - 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}{4 ( s + 1 )}} + L^{- 1} {\frac{1}{4 ( s + 1 ) ^{2}}} + L^{- 1} {\frac{1}{4 ( s - 1 )}} + L^{- 1} {\frac{1}{4 ( s - 1 ) ^{2}}}

L^{- 1} {\frac{1}{4 ( s + 1 )}} : \frac{1}{4} e^{- t}

L^{- 1} {\frac{1}{4 ( s + 1 ) ^{2}}} : \frac{e ^{- t} t}{4}

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

L^{- 1} {\frac{1}{4 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{4}

= - \frac{1}{4} e^{- t} + \frac{e ^{- t} t}{4} + \frac{1}{4} e^{t} + \frac{e ^{t} t}{4}

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

n = 0 \sum \infty \frac{6}{n ( n + 3 )}

f (t) = \frac{1}{t ^{2}}

f (x) = \frac{1}{1 + e ^{x}}, at x = 1

\int \frac{6 - 2 x}{8 - 4 x - 4 x ^{2}} d x

What is inverse oflaplace (s^2)/((s^2-1)^2) ?
The answer to inverse oflaplace (s^2)/((s^2-1)^2) is -1/4 e^{-t}+(e^{-t}t)/4+1/4 e^t+(e^tt)/4