What is inverse oflaplace x/((x^2+1)(x^2+9)) ?

The answer to inverse oflaplace x/((x^2+1)(x^2+9)) is 1/8 cos(t)-1/8 cos(3t)

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inverse oflaplace x/((x^2+1)(x^2+9))

inverse laplace

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

\frac{1}{8} cos (t) - \frac{1}{8} cos (3 t)

Solution steps

L^{- 1} {\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}}

Take the partial fraction of

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )} : \frac{x}{8 ( x ^{2} + 1 )} - \frac{x}{8 ( x ^{2} + 9 )}

= L^{- 1} {\frac{x}{8 ( x ^{2} + 1 )} - \frac{x}{8 ( x ^{2} + 9 )}}

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{x}{8 ( x ^{2} + 1 )}} - L^{- 1} {\frac{x}{8 ( x ^{2} + 9 )}}

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

L^{- 1} {\frac{x}{8 ( x ^{2} + 9 )}} : \frac{1}{8} cos (3 t)

= \frac{1}{8} cos (t) - \frac{1}{8} cos (3 t)

\int t (t^{2} + 1)^{8} d t

\int_{1}^{e^{3}} \frac{ln ^{3} ( x ^{2} )}{x} d x

\frac{x}{1 - x ^{2}}

x \to 0 lim (\frac{( 1 0 ^{x} - 1 )}{x})

x \to 1 + lim (\frac{4}{( x - 1 ) ^{3}})

What is inverse oflaplace x/((x^2+1)(x^2+9)) ?
The answer to inverse oflaplace x/((x^2+1)(x^2+9)) is 1/8 cos(t)-1/8 cos(3t)