determine the inverse Laplace transform of the function. L^{-1}\{R(s)\}=L^{-1}\{\frac{7}{(s+3)(s-3)}\}

Step 1
Given function is,
${\displaystyle R\left(s\right)=\frac{7}{(s+3)(s-3)}}$
This can be re-written as,
${\displaystyle R\left(s\right)=\frac{7}{6}[\frac{1}{s-3}-\frac{1}{s+3}]}$
Step 2
Inverse Laplace is given by,
$L^{-1}\left\{\frac{1}{s+a}\right\}=e^{-at}\text{ and }{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}$
Step 3
Hence, Laplace inverse of the given function is,
$L^{-1}\{R(s)\}=\frac{7}{6}[L^{-1}\left\{\left[\frac{1}{s-3}\right]\right\}-L^{-1}\left\{\frac{1}{s+3}\right\}]$
${\displaystyle =\frac{7}{6}(e^{3t}-e^{-3t})}$