Show that $ \int_{-\infty}^{\infty} \frac{x}{x^4+1} e^{-ikx}dx=- \pi ie^{-k/ \sqrt{2}}\sin(k/ \sqrt{2}) $ for $k>0$

Question

So since $k>0$ we consider the poles in the lower half plane, which are $e^{-i\pi/4}$ and $e^{-3i\pi/4}$

The residue for the first one is $\frac{e^{-i\pi/4}e^{-ikx}}{4e^{-3\pi i/4}}=\frac{1}{4}e^{(\pi/2-kx)i}$

The residue for the second one is $\frac{e^{-3i\pi/4}e^{-ikx}}{4e^{-9\pi i/4}}=\frac{1}{4}e^{(3\pi/2-kx)i}$.

So I get that the integral = $-\frac{2\pi i}{4}(e^{(\pi/2-kx)i}+e^{(3\pi/2-kx)i})$

At this point I'm stuck on where to go from here, since the powers of $e$ are not the same so I don't know how we can get the sin bit?