inverse Laplace Transform: $ L^{-1} \{\log \frac{s^2 - a^2}{s^2} \}$.
Solution 1:
If
$$F(s)=\mathcal{L}\{f(t)\}(s)=\log(1-s^2/a^2)$$
then
$$\mathcal{L}\{t f(t)\}=-F'(s)=-\frac{d}{ds}\log(1-a^2/s^2)=\frac{2}{s}-\frac{1}{s+a}-\frac{1}{s-a}.$$
Now, can you apply the inverse Laplace transform to both sides here? Then just divide by $t$.