The Fourier transform is given by
\begin{equation} \begin{aligned} \hat{f}(\xi) &= \int_{-\infty}^{\infty} f(x) e^{-i\xi x} dx \\ &= \int_{-\infty}^{\infty} \frac{x}{(1+x^2)^2} e^{-i\xi x} dx \end{aligned} \end{equation}
I have a hard time computing this integral. I have tried some substitutions but have gotten stuck every time. After getting stuck I have tried using some Fourier transform rules to simplify the problem. I define
\begin{equation} \begin{aligned} g(x) = \frac{1}{(1+x^2)^2} \end{aligned} \end{equation}
for use with the rule
\begin{equation} \begin{aligned} F[xg(x)] = i(\hat{g})^{'}(\xi) \end{aligned} \end{equation}
To help compute $\hat{g}$ I define
\begin{equation} \begin{aligned} h(x) = \frac{1}{1+x^2} \end{aligned} \end{equation}
for use with the rule
\begin{equation} \begin{aligned} F[f(x)g(x)] = \frac{1}{2\pi} (\hat{f}*\hat{g})(\xi) \end{aligned} \end{equation}
with $f=g$. To compute $\hat{h}$ I have tried using the rule
\begin{equation} \begin{aligned} F \Big[\frac{1}{x^2 + a^2} \Big] = \frac{\pi}{a} e^{-a |\xi|} \end{aligned} \end{equation}
with $a=1$, which gives us
\begin{equation} \begin{aligned} \hat{h}(\xi) &= \pi e^{-|\xi|} \\ F[f(x)g(x)] &= F[h(x)^2] \\ &= \frac{1}{2\pi} (\hat{h}*\hat{h})(\xi) \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{h}(\xi - y) \hat{h}(y) dy \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \pi^2 e^{-|\xi - y|} e^{-|y|} dy \\ &= \frac{\pi}{2} \int_{-\infty}^{\infty} e^{|y||\xi - y|} dy \end{aligned} \end{equation}
This is where I get stuck again, since I have a hard time with this integral as well. I have tried removing the absolute values by considering different intervals, but no luck. I would be grateful for any advice.
Let be $$g(x)=\frac{1}{1+x^2}$$ and $$f(x)=-\frac12g'(x)=\frac{x}{(1+x^2)^2}$$ So we have $$F(\xi)=-\frac12\,i\xi\, G(\xi)$$
By duality theorem $\mathcal F\left\{\frac12\mathrm e^{-|x|}\right\}=\frac{1}{1+\xi^2}$ and then $\mathcal F\left\{\frac{1}{1+x^2}\right\}=\pi\,\mathrm e^{-|\xi|}=G(\xi)$
Hence $$ F(\xi)=-\frac12\,i\xi\, G(\xi)=-\frac12\,i\xi\,\pi\,\mathrm e^{-|\xi|} $$