Inverse laplace transform

Hi, I'm trying to do an inverse laplace transform of the following:

$\displaystyle H(s)=\frac{1}{(1+as)^2}$

I know from a laplace transform table:

$\displaystyle L^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at}$

I'm not sure if this helps me, but I can split the transfer function into partial fractions:

$\displaystyle H(s)=\frac{1}{(1+as)^2}=\frac{A}{(1+as)}+\frac{B}{ (1+as)^2}$

but I'm not sure how I can actually get the time domain function $\displaystyle h(t)=L^{-1}\{H(s))\}$ from these things

Thanks