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? 

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\begin{align}
&\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{st} \over \pars{s^{2} + a^{2}}^{2}}\,{\dd s \over 2\pi\ic} =
\lim_{s \to -a\ic}\partiald{}{s}\bracks{\expo{st} \over \pars{s - a\ic}^{2}} +
\lim_{s \to a\ic}\partiald{}{s}\bracks{\expo{st} \over \pars{s + a\ic}^{2}}
\\[5mm] = &\
2\,\Re\bracks{-\,{-\ic\expo{-\ic at}  + at\expo{-\ic at} \over 4a^{3}}} =
\bbx{\ds{{\sin\pars{at} - at\cos\pars{at} \over 2a^{3}}}}
\end{align}
