Find the inverse of Laplace transform frac{2s+3}{(s-7)^4}

Step 1
Find
$L^{-1}\left\{\frac{2s+3}{(s-7{)}^4}\right\}$
For this take partial fraction
$\frac{2s+3}{(s-7{)}^4}=\frac{2}{(s-7{)}^3}+\frac{17}{(s-7{)}^4}$
Step 2
Use linearity property of Inverse Laplace transformation
That is , for function f(s),g(s) and a,b be any constants.
$L^{-1}\{af(s)+bf(s)\}=a{L}^{-1}\{f(s)\}+b{L}^{-1}\{g(s)\}$
Step 3
Now equation (i) becomes
$L^{-1}\left\{\frac{2s+3}{(s-7{)}^4}\right\}=L^{-1}\left\{\frac{2}{(s-7{)}^3}\right\}+L^{-1}\left\{\frac{17}{(s-7{)}^4}\right\}=e^{7t}{t}^2+\frac{17e^{7t}{t}^3}{6}$
Formula used
$L^{-1}(F(s))=f(t)$
$L^{-1}(F(S-a))=e^{at}f(t)$
$L^{-1}\left(\frac{n!}{S^{n+1}}\right)=t^n$