On the Fourier transform of $f(x)=\ln(x^2+a^2)$ I would like to derive the Fourier transform of $f(x)=\ln(x^2+a^2)$, where $a\in \mathbb{R}^+$ by making use of the properties:
\begin{equation}
\mathcal{F}[f'(x)]=(ik)\hat{f}(k)\\
\mathcal{F}[-ixf(x)]=\hat{f}'(k)
\end{equation}
For the Fourier transform I use the definition given by:
\begin{equation}
\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, k \in \mathbb{R}
\end{equation}
Until now I found out that by taking the derivative of $f$ and finding the Fourier transform of $f'$ I can then use the relation $\mathcal{F}[f'(x)]=(ik)\hat{f}(k)$ and find $\hat{f}$. The derivative of $f$ would be:
\begin{equation}
f'(x)=\frac{2x}{x^2+a^2}
\end{equation}
and by considering $g(x)=1/(x^2+a^2)$, I then have:
\begin{equation}
f'(x)=2xg(x)
\end{equation}
Now I know that the Fourier transform of $g$ is given by:
\begin{equation}
\hat{g}(k)=\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|k|}, a \in \mathbb{R}, k\in \mathbb{R}
\end{equation}
Now I must find the Fourier transform of $xg(x)$ which would be given by the derivative of $\hat{g}$ right? But how can this possible since $\hat{g}$ has no derivative? 
I think I am really close now but I need that extra tip.
Thank you!

If $1/|x|$ means even and homog degree $-1$, the only such distn is $\delta$. Attempting to alter this fact by arbitrary truncations can indeed produce several different seeming answers. That’s not good.
I’d prefer to continue the approach in the question: up to the constant there, the desired Fourier transform is the derivative of $e^{-|ak|}$ in $k$. Without loss of generality, $a>0$. Then the deriv is $-sgn(x)ae^{-|ak|}-2a\delta$. Then mult by the ${1\over a}\sqrt{\pi/2}$.
