Laplace Inverse of tan inverse

Question

Obtain inverse Laplace transform of $$\tan^{-1}\left(\frac{2}{s^{2}}\right).$$

Here is my work so far: \begin{align*} L^{-1}\left[\tan^{-1}\left(\frac{2}{s^{2}}\right)\right]&=f(t)\\ L^{-1}\left[\frac{d}{ds}\left(\tan^{-1}\left(\frac{2}{s^{2}}\right)\right)\right]&=-tf(t)\\ L^{-1}\left[ \frac{1}{1+\left(\frac{2}{s^{2}}\right)^{2}}\cdot\left(\frac{-4}{s^{3}}\right)\right]&=-tf(t)\\ L^{-1}\left[\frac{4s}{s^{4}+4}\right]&=tf(t). \end{align*}

I am unable to proceed after this step.

Answer 1

As Kaynex suggested, noting $$ s^4+4=(s^4+4s^2+4)-4s^2=(s^2-2s+2)(s^2+2s+2) $$ and $$ \frac{4s}{s^4+4}=\frac{1}{s^2-2s+2}-\frac{1}{s^2+2s+2}=\frac{1}{(s-1)^2+1}-\frac{1}{(s+1)^2+1} $$ so one has \begin{eqnarray} L^{-1}\bigg\{\frac{4s}{s^4+4}\bigg\}&=&L^{-1}\bigg\{\frac{1}{(s-1)^2+1}\bigg\}-L^{-1}\bigg\{\frac{1}{(s+1)^2+1}\bigg\}\\ &=&(e^t-e^{-t})\sin t. \end{eqnarray}