How do we find the inverse Laplace transform of $\frac{1}{(s^2+a^2)^2}$? How do we find the inverse Laplace transform of $\frac{1}{(s^2+a^2)^2}$? Do I need to use the convolution theory? It doesn't match any of the known laplace inverse transforms. It matches with the Laplace transform of $\sin(at)$ but I don't know if that helps or not. Also there seems to be a formula with limits and imaginary numbers. How do I just know to apply that here? 

Hint: Use $$\frac{d}{ds}\left(\frac{s}{s^2+a^2}\right)=2a^2\frac{1}{(s^2+a^2)^2}-\frac{1}{s^2+a^2}$$
