Fourier transform of $f(x)=\frac{x}{1+x^4}$ and $g(x)=\frac{x^2}{1+x^4}=xf(x)$

Question

Let $f(x)=x/(1+x^4)$, the improper integral of which exists.

I computed the Fourier transform of $f$, to be:

\begin{equation} \mathcal{F}(f(x))=\hat{f}(k)=-\sqrt{\frac{\pi}{2}}\left(H(-k)ie^{(k/\sqrt{2})}sin(k/\sqrt{2})+H(k)e^{-(ik/\sqrt{2})}sinh(k/\sqrt{2}) \right), k \in \mathbb{R} \end{equation} where: \begin{equation} H(k)=\begin{cases} 0, & k<0 \\ 1, & k\geq 0\end{cases} \end{equation}

• My first question is whether this is correct or not.
• My second and last one, has to do with the computation of the Fourier transform for $g(x)=x^2/(1+x^4)$. I am pretty sure that I do not have to go through the whole procedure again in order to compute this one. Actually, I think that I could just as well compute the transform for $h(x)=1/(1+x^4)$ and then derive the rest from this one.

But, how?

Thank you!