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 $\left| x \right|^{-1}$ is the distribution defined as
$$\left( \left| x \right|^{-1}, \phi \right) =
\int_{\left| x \right| < 1} \frac {\phi(x) - \phi(0)} {\left| x \right|} dx +
 \int_{\left| x \right| > 1} \frac {\phi(x)} {\left| x \right|} dx,$$
then
$${\mathcal F}\!\left[ \ln\left( x^2 +a^2 \right) \right] =
\frac 1 {\sqrt{2 \pi}}
 \int_{-\infty}^{\infty} \ln\left( x^2 +a^2 \right) e^{-i k x} dx =
-\sqrt{2 \pi} \left( \frac { e^{-a \left| k \right|}} {\left| k \right|} +
 2 \gamma \delta(k) \right),$$
where $\gamma$ is Euler's constant. Probably the easiest way to prove it is to compute the inverse transform directly from the definition of $\left| x \right|^{-1}$.
