The derivative of $\log(a^2+s^2)$ is $\frac{2s}{a^2+s^2}$. Now, we know that the fourier transform of $f'(x)$ is $i\omega \hat{f}(\omega)$. So we can find the fourier transform of $\frac{2s}{a^2+s^2}$ and then divide it by $i\omega$ to find our answer. We can use the Residue Theorem to evaluate $\int_{-\infty}^{\infty} \frac{2s}{a^2+s^2}e^{-i\omega s}ds$.
There are poles at $\pm ia$. Depending on the sign of $\omega$, we either do our contour integral in the upper half or lower half of the plane. Let's do it in the lower half plane. The only residue in our domain is at $-ia$. Our residue is :$$\frac{2s}{2s}e^{-i \omega s}=e^{-i \omega s}$$ For $s=-ia$, we get $e^{-\omega a}$
With the Residue Theorem, we get $-2 \pi i e^{-\omega a}$ (minus because we are in the lower half plane), which is the Fourier Transform of the derivative. Now we divide by $i \omega$ to find the FT of the original function, and we get : $$\frac{-2 \pi}{\omega}e^{- \omega a}$$ which we could rewrite as ($\omega= 2 \pi \xi $ , $\xi$ is the frequency) $$\frac{-1}{\xi}e^{- \omega a}$$
Hope this helps !
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Note 2) : this is my first answer. If I did any mistake, please say it in the comments and I will correct my answer. Thank you !