Inverse Laplace Transformation requiring a lot of factoring

Question

I was not able to factor and therefore don't know where to start, any possible hint?

Answer 1

Hint : $$\frac{as+b}{s^2+6s+13}+\frac{cs+d}{(s^2+6s+13)^2}=\frac{as^3+(6a+b)s^2+(13a+6b+c)s+13b+d}{(s^2+6s+13)^2}$$

Answer 2

Hint: just cheat. The Laplace transforms of the given options are simple to compute. And since $\frac{s^3+7s^2+15s+1}{(s^2+6s+13)^2}$ at $s=0$ equals $\frac{1}{169}$, that has to be the integral of the correct option over $(0,+\infty)$.

It follows that if there is a correct option, it is the first one.