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

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

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

inverse laplace

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

\frac{t cos ( t ) + sin ( t )}{2}

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

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

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

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

= sin (t) - \frac{1}{2} (sin (t) - t cos (t))

Simplify

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

= \frac{t cos ( t ) + sin ( t )}{2}

ln (\frac{( x + 6 ) ^{4} e ^{5 x^{3} - 3}}{x + 2 x ^{3}})

y = x^{2} + 4 x + 7, at x = 1

\frac{7}{2 7 x + 1}, at x = 9

ln (y^{2})

\frac{d}{d x} (arctan (1 - x))

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