How to quickly compute the inverse laplace transform of $\frac{1}{(s^2+1)^2}$ I wish to find the inverse laplace transform of $\dfrac{1}{(s^2+1)^2}$
I tried using partial fraction but things do not seem to work out i.e.
$\dfrac{1}{(s^2+1)^2} = \dfrac{A}{s^2+1} + \dfrac{B}{s^2+1}$
$\Rightarrow$ $As^2 + Bs^2 = 0$, $A + B = 1$ which is inconclusive
Is there a quick way to find the inverse laplace transform of this guy?

I guess you know the rule
$$ \mathcal{L}^{-1}[ f'(s) ](t) =- t \,\mathcal{L}^{-1} [f(s)](t)$$
and the fact that
$$\mathcal{L}^{-1} \Bigl[\frac{1}{s^2+1} \Bigr](t) = \sin(t)$$
and
$$\mathcal{L}^{-1} \Bigl[\frac{s}{s^2+1} \Bigr](t) = \cos(t)\quad ?$$
Now observe that
$$\frac{d}{ds} \biggl(\frac{s}{s^2+1}\biggr) = -\frac{1}{(s^2+1)} +\frac{2}{(s^2+1)^2}$$
and thus
$$\frac{1}{(s^2+1)^2}= \frac12\frac{d}{ds} \biggl(\frac{s}{s^2+1}\biggr)  + \frac12 \frac{1}{(s^2+1)}$$
Using these results, we obtain
$$\mathcal{L}^{-1}\biggl[ \frac{1}{(s^2+1)^2} \biggr](t)
= -\frac{1}{2} t\,\mathcal{L}^{-1} \biggl[ \frac{s}{s^2+1}\biggr] (t)+\frac{1}{2} \sin t= \frac{-t \cos t +\sin t}{2} $$
