# Please provide steps The inverse Laplace transform for F(s)=8/(s+9)-6/(s^2-sqrt3)

The inverse Laplace transform for
$F\left(s\right)=\frac{8}{s+9}-\frac{6}{{s}^{2}-\sqrt{3}}$ is
a) $8{e}^{-9t}-6\mathrm{sin}h\left(3t\right)$
b) $8{e}^{-9t}-6\mathrm{cos}h\left(3t\right)$
c) $8{e}^{9t}-6\mathrm{sin}h\left(3t\right)$
d) $8{e}^{9t}-6\mathrm{cos}h\left(3t\right)$

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Step 1
Consider the function:
$F\left(s\right)=\frac{8}{x+9}-\frac{6}{{x}^{2}-\sqrt{3}}$
Find Inverse Laplace Transform.
Apply Partial fraction on $\frac{6}{{x}^{2}-\sqrt{3}}$
$\frac{6}{{x}^{2}-\sqrt{3}}=\frac{A}{x-\sqrt[4]{3}}+\frac{B}{x+\sqrt[4]{3}}$
$6=A\left(x+\sqrt[4]{3}\right)+B\left(x-\sqrt[4]{3}\right)$
Let and find A and B
$A=\frac{3}{\sqrt[4]{3}}$
$B=-\frac{3}{\sqrt[4]{3}}$
$F\left(s\right)=\frac{8}{x+9}-\frac{3}{\sqrt[4]{3}\left(x-\sqrt[4]{3}\right)}+\frac{3}{\sqrt[4]{3}\left(x+\sqrt[4]{3}\right)}$
Step 2
Inverse Laplace transform property

${L}^{-1}\left\{\frac{8}{x+9}-\frac{3}{\sqrt[4]{3}\left(x-\sqrt[4]{3}\right)}+\frac{3}{\sqrt[4]{3}\left(x+\sqrt[4]{3}\right)}\right\}=8{L}^{-1}\left\{\frac{1}{x+9}\right\}-{3}^{\frac{3}{4}}{L}^{-1}\left\{\frac{1}{x-\sqrt[4]{3}}\right\}+{3}^{\frac{3}{4}}{L}^{-1}\left\{\frac{1}{x+\sqrt[4]{3}}\right\}$
$=8{e}^{-9t}-{3}^{\frac{3}{4}}{e}^{\sqrt[4]{3}t}+{3}^{\frac{3}{4}}{e}^{-\sqrt[4]{3}t}$
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