inverse Laplace transform of 1/s^5 - (5 - 9s) / (s^2 + 100)

Find the inverse Laplace transforms of the function given

\(\displaystyle \L\\ 1/s^5 - (5-9s)/(s^2 + 100)\)

I get

\(\displaystyle \L\\ (t^4)/24\)

but I don’t know what to do for the second half


the only thing I can come up with is

\(\displaystyle \L\\ [s^2 + 100] = [s^2 + 10^2]\)

but I don’t think that helps much.

Expand and you'll see a more recognizable form for the two parts.


\(\displaystyle \L\\\frac{5-9s}{s^{2}+100}=\frac{5}{s^{2}+100}-\frac{9s}{s^{2}+100}\)


\(\displaystyle \L\\9cos(10t)-\frac{sin(10t)}{2}+\frac{t^{4}}{24}\)

I don't know why I didn't see that before. Thank you!