# Thread: help finding inverse laplace

1. ## help finding inverse laplace

1/((s^2)+1)^3

2. Originally Posted by self_shattered

1/((s^2)+1)^3
Get a decent table of Laplace transforms and look it up:

$\frac{1}{s^2+a^2}=\mathcal{L}\left( \frac{(3-a^2t^2)\sin(at)-3at \cos(at)}{8a^5} \right)$

CB

3. Originally Posted by self_shattered
$\mathbb{L}^{-1}\left\{\frac{1}{(s^2+1)^3}\right\}=1/8\left(3\sin(t)-3t\cos(t)-t^2\sin(t)\right)$