I thought it might be instructive to present an approach that develops the Fourier Transform directly and is distinct and complements the approach I adopted in THIS ANSWER. To that end, we now proceed.
Let $f(x)=\log(x^2+a^2)$ and let $F(k)$ denote its Fourier Transform. Then, we write
$$F(k)=\mathscr{F}\{f\}(k)\tag 1$$
where $(1)$ is interpreted as a Tempered Distribution. That is, for any $\phi \in \mathbb{S}$, we can write for any $\nu>0$
$$\begin{align}
\langle \mathscr{F}\{f\}, \phi\rangle &=\langle f, \mathscr{F}\{\phi\}\rangle\\\\
&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \log(x^2+a^2)\int_{-\infty}^\infty \phi(k)e^{-ikx}\,dk\,dx\\\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty \log(x^2+a^2)\int_{-\infty}^\infty \phi(k)\cos(kx)\,dk\,dx\\\\
&=\frac4{\sqrt{2\pi}}\phi(0)\int_0^\infty \frac{\sin(\nu x)}{x}\,\log(x^2+a^2)\,dx\\\\
&+\frac2{\sqrt{2\pi}}\int_0^\infty \log(x^2+a^2) \left(\int_{-\infty}^\infty (\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k))\cos(kx)\,dk\right)\,dx\\\\
&=C_\nu(a) \phi(0)\tag2\\\\
&+\frac2{\sqrt{2\pi}}\lim_{L\to \infty } \int_{-\infty}^\infty \frac{\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)}k\left(\log(L^2+a^2)\sin(kL)\\-\int_0^L \frac{2x\sin(kx)}{x^2+a^2}\,dx\right)\,dk\\\\
&=C_\nu(a) \phi(0)-\sqrt{2\pi}\int_{-\infty}^\infty \left(\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)\right)\frac{e^{-|ka|}}{|k|}\,dk\tag3
\end{align}$$
where the constant $C_\nu(a)$ is given by
$$\begin{align}
C_\nu(a)&=-2\sqrt{2\pi} \log(\nu)+\frac4{\sqrt{2\pi}}\int_0^\infty \frac{\sin(x)}{x}\,\log(x^2+\nu^2a^2)\,dx\\\\
&=2\sqrt{2\pi} (\log(a)-\text{Ei}(-\nu a))
\end{align}$$
NOTES:
In arriving at $(2)$, used Fubini's theorem to justify interchanging integral $\int_0^L \,dx$ with the integral $\int_{-\infty}^\infty \,dk$.
In arriving at $(3)$, we used integration by parts to show that the limit of the term involving $\log(L^2+a^2)$ is $0$ and we appealed the the Dominated Convergence Theorem to justify interchanging the limit with the integration over $k$ for the second term on the right-hand side of $(3)$.
For $\nu=1$ and $a\to 0$, $C_1(a\to 0)=-2\sqrt{2\pi}\gamma$, which agrees with the results in the previously referenced answer (modulo the scale factor $1/\sqrt{2\pi}$ here from the definition of the Fourier Transform).
From $(3)$, we deduce the Fourier Transform of $\psi$ in distribution
$$\bbox[5px,border:2px solid #C0A000] {\mathscr{F}\{f\}(k)=C_\nu(a) \delta(k)-\sqrt{2\pi}\left(\frac{e^{-|ka|}}{|k|}\right)_\nu}\tag4$$
where the distribution $\left(\frac{e^{-|ka|}}{|k|}\right)_\nu$ is defined by its action on any $\phi\in \mathbb{S}$
$$\int_{-\infty}^\infty \left(\frac{e^{-|ka|}}{|k|}\right)_\nu\phi(k)\,dk=\int_{-\infty}^\infty \left(\phi(k)-\phi(0)\xi_{[-\nu,\nu]}(k)\right)\frac{e^{-|ka|}}{|k|}\,dk$$
SOLUTION VERIFICATION:
In this section, we take the inverse Fourier Transform of the distribution in $(6)$ and show that it is indeed equal to $\log(x^2+a^2)$. To begin, we write for any $\phi\in \mathbb{S}$
$$\begin{align}
\langle \mathscr{F}^{-1}\{F\},\phi\rangle&=\langle F,\mathscr{F}^{-1}\{\phi\}\rangle \\\\
&=C_\nu(a)\mathscr{F}^{-1}\{\phi\}(0)-\int_{-\infty}^\infty \frac{e^{-|ka|}}{|k|}\int_{-\infty}^\infty \phi(x)\left(e^{ikx}-\xi_{[-\nu,\nu]}(k)\right)\,dx\,dk\\\\
&=C_\nu(a)\mathscr{F}^{-1}\{\phi\}(0)\\\\
&-2\int_{-\infty}^\infty \phi(x)\int_{0}^\infty \frac{e^{-ka}}{k}\left(\cos(kx)-\xi_{[-\nu,\nu]}(k)\right)\,dk\,dx\tag5
\end{align}$$
Using "Feynman's Trick," we find that the inner integral on the right-hand side of $(5)$ is given by
$$\begin{align}
\int_{0}^\infty \frac{e^{-ka}}{k}\left(\cos(kx)-\xi_{[-\nu,\nu]}(k)\right)\,dk&=\int_\infty^a \left(\frac{1-e^{-\nu a'}}{a'}-\frac{a'}{x^2+a'^2}\right)\,da'\\\\
&=\log(a)+\int_{\nu a}^\infty \frac{e^{-a' }}{a'}\,da'-\frac12\log(x^2+a^2))\\\\
&=-\frac12\log(x^2+a^2)+\log(a)+\text{Ei}(-\nu a)\tag6
\end{align}$$
Finally, substituting $(6)$ in $(5)$ yields the anticipated result that
$$\langle \mathscr{F}^{-1}\{F\},\phi\rangle=\int_{-\infty}^\infty \phi(x)\log(x^2+a^2)\,dx$$
from which we conclude that
$$f(x)=\mathscr{F}^{-1}\{F\}(x)=\log(x^2+a^2)$$
