$ X(s) = \frac{s}{s^4+1} $
I tried to use the definition: $ f(t) = \mathcal{L}^{-1} \{F(s)\} = \frac{1}{2 \pi i} \lim_{T\to\infty}\int_{ \gamma - i T}^{ \gamma + i T} e^{st} F(s)\,ds $
or the partial fraction expansion but I have not achieved results.
Hint: With $$s^4+1=s^4+1+2s^2-2s^2=(s^2+1-\sqrt{2}s)(s^2+1+\sqrt{2}s)$$ then $$\dfrac{s}{s^4+1}=\dfrac{\sqrt{2}}{4}\dfrac{1}{s^2+1-\sqrt{2}s}-\dfrac{\sqrt{2}}{4}\dfrac{1}{s^2+1-\sqrt{2}s}$$ then complete the squares in denominators!