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

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

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

inverse laplace

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

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

Solution steps

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

= L^{- 1} {\frac{1}{2} \cdot \frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{2} L^{- 1} {\frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{2 a ^{3}}{( s ^{2} + a ^{2} ) ^{2}}} = sin (a t) - a t cos (a t)

L^{- 1} {\frac{2 \cdot 1 ^{3}}{( x ^{2} + 1 ^{2} ) ^{2}}} = sin (t) - t cos (t)

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

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

\int_{0}^{1} x y d x

\frac{d}{d x} ((2 x^{2} + 5) (4 x - 1))

y^{^{''}} - (\frac{3}{x}) y^{^{'}} - (\frac{6}{x ^{2}}) y = 0

ln (5 + e^{x y}) = y

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