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

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

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

inverse laplace

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

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

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

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

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 1 ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 1 ) ^{2}}} (0)

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

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

Evaluate

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

at point

0 = 0

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

(\frac{( 3 x ^{2} )}{( 2 x + 4 )})^{3}

\frac{d}{d x} ((x - x) (x + x))

\int (3 x - 4) e - x d x

\frac{\partial}{\partial x} (\frac{y ^{2}}{2})

\frac{d}{d x} (\frac{- 9 x - 1}{- 6 x + 2})

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