Find the inverse Laplace transform of: (9s+6)/((s^2)-49)

I separate this into two fractions: 9s/(s^2-49) + 6/(s^2-49)

For the first fraction, I think I am correct in using the rule that for a transform of s/(s^2-a^2) ...the inverse transform is coshat. And for the second fraction, the inverse transform of a/(s^2-a^2) is sinhat. I'm not sure what the "h" means or how to use it.

Thanks =)