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

The answer to inverse oflaplace-1/2*1/((x^2+1)^2) is -1/4 (sin(t)-tcos(t))

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

inverse laplace

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

- \frac{1}{4} (sin (t) - t cos (t))

Solution steps

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

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

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

Refine

- \frac{1}{2} \frac{1}{2} (sin (t) - t cos (t)) : - \frac{1}{4} (sin (t) - t cos (t))

= - \frac{1}{4} (sin (t) - t cos (t))

\int_{0}^{2} (2 x - x^{2})^{2} d x

\int_{5}^{6} \frac{x}{x - 5} d x

\int \frac{x ^{2}}{x + 5} d x

x = 9 y^{3}, x = 9 y^{2}

\frac{d}{d x} (sin (x) ln (3 x))

What is inverse oflaplace-1/2*1/((x^2+1)^2) ?
The answer to inverse oflaplace-1/2*1/((x^2+1)^2) is -1/4 (sin(t)-tcos(t))