# determine the inverse Laplace transform of the given function. F(s)=\frac{2s}{s^2+9}

determine the inverse Laplace transform of the given function.
$F\left(s\right)=\frac{2s}{{s}^{2}+9}$
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Step 1
Given
$F\left(s\right)=\frac{2s}{{s}^{2}+9}$
Step 2
${L}^{-1}\left\{\frac{s}{{s}^{2}+{a}^{2}}\right\}=\mathrm{cos}at$
so
${L}^{-1}\left\{\frac{2s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+{3}^{2}}\right\}$
$=2\mathrm{cos}3t$

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