What is inverse oflaplace (2x)/((x^2+4)^2) ?

The answer to inverse oflaplace (2x)/((x^2+4)^2) is (tsin(2t))/2

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inverse oflaplace (2x)/((x^2+4)^2)

inverse laplace

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

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

Solution steps

L^{- 1} {\frac{2 x}{( x ^{2} + 4 ) ^{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{2 x}{( x ^{2} + 4 ) ^{2}} : F (x) = \frac{2}{( x ^{2} + 4 ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{2}{( x ^{2} + 4 ) ^{2}}}) + L^{- 1} {\frac{2}{( x ^{2} + 4 ) ^{2}}} (0)

L^{- 1} {\frac{2}{( x ^{2} + 4 ) ^{2}}} : \frac{1}{8} (sin (2 t) - 2 t cos (2 t))

\frac{d}{d t} (\frac{1}{8} (sin (2 t) - 2 t cos (2 t))) = \frac{t sin ( 2 t )}{2}

Evaluate

\frac{1}{8} (sin (2 t) - 2 t cos (2 t))

at point

0 = 0

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

\frac{d}{d x} (ln (arctan (8 x^{5})))

\frac{d}{d x} (x^{3} (x - 2)^{2})

\int 3 x \cdot (x + 1)^{2} d x

(1 + y^{2} + x y^{2}) d x + (x^{2} y + y + 2 x y) d y = 0

\frac{d}{d x} (x^{0.6})

What is inverse oflaplace (2x)/((x^2+4)^2) ?
The answer to inverse oflaplace (2x)/((x^2+4)^2) is (tsin(2t))/2