# Inverse Laplace transform question help $\frac{1}{(s^2+1)^2} - \frac{1}{s^2(s^2+1)^2}$

I am having a hard time finding the inverse Laplace transform of $$\frac{1}{(s^2+1)^2} - \frac{1}{s^2(s^2+1)^2}$$ and would appreciate some guidance. I have tried breaking it down to partial fractions but got stuck.

First get rid of the ${ 1\over s^2}$ part as in: $\frac{1}{(s^2+1)^2} - \frac{1}{s^2(s^2+1)^2} = \frac{1}{{s}^{2}+1}+\frac{2}{{( {s}^{2}+1) }^{2}}-\frac{1}{{s}^{2}}$.
Then using $s^2+1 = (s-i)(s+i)$ we write $\frac{1}{{s}^{2}+1}+\frac{2}{{( {s}^{2}+1) }^{2}} = \frac{a}{s+i}+\frac{b}{s-i}+\frac{c}{{( s+i) }^{2}}+\frac{d}{{( s-i) }^{2}}$, and solving gives $a=i, b=-i, c=d = -{1 \over 2}$.