Calculate the Fourier transform of $\log |x| $
One standard way to compute the Fourier Transform of this kind of function is to multiply by $e^{-\epsilon x^2}$ and let $\epsilon\to0$. $$ \begin{align} &\lim_{\epsilon\to0}\int_{-\infty}^\infty e^{-\epsilon x^2}\log\!|x|\,e^{-ix\xi}\,\mathrm{d}x\\ &=2\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\log(x)\,\cos(x|\xi|)\,\mathrm{d}x\\ &=\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\log(x)\,\mathrm{d}\sin(x|\xi|)\\ &=-\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty e^{-\epsilon x^2}\frac{\sin(x|\xi|)}x\,\mathrm{d}x +\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty2\epsilon xe^{-\epsilon x^2}\log(x)\sin(x|\xi|)\,\mathrm{d}x\\ &=-\frac\pi{|\xi|}+\frac2{|\xi|}\lim_{\epsilon\to0}\int_0^\infty2\epsilon xe^{-\epsilon x^2}(\log(x)-\log(|\xi|))\sin(x)\,\mathrm{d}x\\ &=-\frac\pi{|\xi|}\tag{1} \end{align} $$
We can estimate $$ \frac1M\int_0^M\log(x)\sin(x)\,\mathrm{d}x=O\!\left(\frac{\log(M)}M\right)\tag{2} $$ because $$ \begin{align} &2\int_0^{2k\pi}\log(x)\sin(x)\,\mathrm{d}x -\overbrace{\int_0^\pi\log(x)\sin(x)\,\mathrm{d}x}^\text{constant} +\int_{2k\pi}^{(2k+1)\pi}\log(x)\sin(x)\,\mathrm{d}x\\ &=\int_0^{2k\pi}\log(x)\sin(x)\,\mathrm{d}x +\int_\pi^{(2k+1)\pi}\log(x)\sin(x)\,\mathrm{d}x\\ &=\int_0^{2k\pi}(\log(x)-\log(x+\pi))\sin(x)\,\mathrm{d}x\\[9pt] &=O(1)\tag{3} \end{align} $$
Define $$ F(x)=\frac1x\int_0^xf(t)\,\mathrm{d}t\implies f(x)=F(x)+xF'(x)\tag{4} $$ then $$ \begin{align} \int_0^\infty2xe^{-x^2}f(x)\,\mathrm{d}x &=\int_0^\infty2xe^{-x^2}(F(x)+xF'(x))\,\mathrm{d}x\\ &=\int_0^\infty2xe^{-x^2}F(x)\,\mathrm{d}x+\int_0^\infty2x^2e^{-x^2}\,\mathrm{d}F(x)\\ &=\int_0^\infty2xe^{-x^2}F(x)\,\mathrm{d}x-\int_0^\infty F(x)\left(4x-4x^3\right)e^{-x^2}\,\mathrm{d}x\\ &=\int_0^\infty\left(4x^3-2x\right)e^{-x^2}F(x)\,\mathrm{d}x\tag{5} \end{align} $$
Equations $(2)$ and $(5)$ validate the last step in $(1)$.
Although the computation above does not compute the $\delta(x)$ component, since $-\frac\pi{|\xi|}$ is not integrable near $0$, any test function must be $0$ at $0$. This means that the coefficient of $\delta(x)$ is irrelevant.
I will start from the well-known expression (see here) for the Euler-Mescheroni constant $\gamma$:
$$\gamma=\int_0^1\frac{1-\cos t}{t}dt-\int_{1}^\infty\frac{\cos t}{t}dt$$ Now, if for $ x>0$ we consider $$F(x)=\int_0^x\frac{1-\cos t}{t}dt-\int_{x}^\infty\frac{\cos t}{t}dt$$ Then we conclude from $ F(1)=\gamma$ and $F'(x)=1/x$ that $$\eqalign{\gamma+\ln x&= \int_0^x\frac{1-\cos t}{t}dt-\int_{x}^\infty\frac{\cos t}{t}dt\cr &=\int_0^1\frac{1-\cos(xt)}{t}dt-\int_{1}^\infty\frac{\cos (xt)}{t}dt }$$ And since the right side of the above formula is even we conclude that $$ \gamma+\ln|x|=\int_0^\infty\frac{\mathbb{I}_{[0,1]}(t)-\cos(xt)}{t}dt\tag1 $$ For every nonzero $x$.
Let the regular distribution assosiated with the function $x\mapsto \gamma+\ln|x|$ be denoted by $T$. What is the action of $T$ on some test function $\phi$?
Indeed, if $\phi$ is a function from $\mathcal{S}$ then using $(1)$ we see that $$\eqalign{\langle T,\phi\rangle &=\int_{\mathbb{R}}(\gamma+\ln|x|)\phi(x)dx\cr &=\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\hat{\phi}(0)-\hat{\phi}(t)-\hat{\phi}(-t)}{2t}dt\cr &=\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\hat{\phi}(0)-\hat{\phi}(t)-\check{\hat{\phi}}(t)}{2t}dt\tag2 }$$ Indeed, since $\hat{\phi}(t)=\int_{\mathbb{R}}\phi(x)e^{-ixt}dx$ we see easily that $$\hat{\phi}(0)=\int_{\mathbb{R}}\phi(x)dx\quad\hbox{and}\quad \hat{\phi}(t)+\hat{\phi}(-t)=2\int_{\mathbb{R}}\phi(x)\cos(xt)dx$$ Thus, applying (2) to $\hat {\phi}$ and noting that $\hat{\hat{\phi}}=2\pi\check{\phi}$ we conclude that $$\eqalign{\langle T,\hat{\phi}\rangle &=2\pi\int_0^\infty\frac{2\mathbb{I}_{[0,1]}(t)\phi(0)-\phi(t)-\phi(-t)}{2t}dt\cr &=\pi\int_0^1\frac{2\phi(0)-\phi(t)-\phi(-t)}{t}dt-\pi \int_1^\infty\frac{\phi(t)+\phi(-t)}{t}dt\cr &=\pi\int_{0}^\infty (\ln t)(\phi'(t)-\phi'(-t))dt\qquad\hbox{(integration by parts)}\cr &=\pi\int_{\mathbb{R}}{\rm sign}(t)\ln|t|\phi'(t)dt\cr &=-\pi\langle{\rm pf}\frac{1}{|x|},\phi\rangle }$$ So, $\hat{T}=-\pi\,{\rm pf}\frac{1}{|x|}$. But $$\widehat{(\gamma+\ln|x|)}=2\pi\gamma\delta+\widehat{\ln|x|}$$ Thus $$\widehat{\ln|x|}=-\pi\,{\rm pf}\frac{1}{|x|}-2\pi\gamma\delta$$ Which is the desired conclusion.
Yes, such computations are standard, but/and can be done in several ways. One approach is to first observe that ${d\over dx}(\log|x|)$ is the principal-value integral $u$ against $1/x$ (not $1/|x|$). This principal value integral is not a literal integrate-against functional, since $1/x$ is not locally integrable (nor is $1/|x|$). Even though it's not a literal integral, one still shows directly that $x\cdot u = 1$, where on the left multiplication by the smooth function (of moderate growth...) on tempered distributions is as usual. (We simply cannot "divide" in a pointwise sense.)
Fourier transform has an easily-verified effect on positive-homogeneity, and parity: the FT of $|x|^{-s}$ is a constant multiple of $|x|^{1-s}$, literally so for $0<\Re(s)<1$, and then by meromorphic continuation. Thus, the Fourier transform of $u$ is a constant multiple of $\mathrm{sgn}\,x$. By integrating against $xe^{-\pi x^2}$, for example (or almost any other odd Schwartz function) one finds that the constant is $-i\pi$ (maybe!).
Thus, letting $F$ be Fourier transform, $$ -i\pi \mathrm{sgn}\,x \;=\; F u \;=\; F{d\over dx}\log|\cdot| \;=\; -2\pi ix \cdot F\log|\cdot| $$ Thus, $2x\cdot F\log|\cdot|= \mathrm{sgn}\,x$. Again, we cannot quite divide pointwise. However, the kernel of the multiplication-by-$x$ operator on tempered distributions consists of distributions supported at $\{0\}$, which (essentially by the theory of Taylor-Maclaurin series) is just finite linear combinations of Dirac $\delta$ and its derivatives. Further, the only such linear combination annihilated by mult'n by $x$ are just multiples of $\delta$ itself. Thus, the relation $2x\cdot F\log|\cdot|=\mathrm{sgn}\,x$ determines that Fourier transform up to multiples of $\delta$.
To determine the constant, let $g(x)=e^{-\pi x^2}$, for example, and for arbitrary Schwartz function $f$, use the standard trick $$ v(f) \;=\; v(f-f(0)\cdot g)+f(0)v(g) $$ and then evaluate $v(f-f(0)g)$ by using the literal integral definition, since $f-f(0)g$ vanishes at $0$, etc.
EDIT: per request of the questioner, I'll give the determination-of-constant idea (in principle standard, but... etc) in further detail, though I would have to think more to express it in terms of the Euler-Mascheroni constant, etc.
That is, let $u=\widehat \log|\cdot|$. Suppose we know that $x\cdot u=a\cdot \mathrm{sgn}\,x$, where I've written another constant $a$ to accommodate possible earlier boo-boos, and make it easier to track. Also note that for a test function $f$, if $f(0)=0$, then $f(x)/x$ is also a test function. Let $g$ be the Gaussian, as above. Then $f(x)-f(0)\cdot g(x)$ is of the form $x\cdot h(x)$ for a test function $h$. Thus, $$ u(f) \;=\; u(f-f(0)g)+f(0)u(g) \;=\; u(x\cdot {f-f(0)g\over x}) + \delta f \cdot u(g) \;=\; (x\cdot u)({f-f(0)g\over x}) + \delta f\cdot u(g) $$ $$ \;=\; a\int \mathrm{sgn}\,(x)\cdot {f(x)-f(0)g(x)\over x}\;dx + \delta f\cdot u(g) \;=\; a\int {f(x)-f(0)g(x)\over |x|}\;dx + \delta f\cdot u(g) $$ The integral can be further explicated in various ways, e.g., integrating by parts. The most-unknown part of the business is the constant $u(g)$, which appears (maybe part of) the coefficient of $\delta$.
Edit-Edit: in response to some further questions: to see the vanishing at $0$ of $f-f(0)g$: $$ f(0)-f(0)\cdot g(0) \;=\; f(0) - f(0)\cdot 1 \;=\; 0 $$ The fact that $$ u(f) \;=\; u(f-f(0)g+f(0)g)\;=\;u(f-f(0)\cdot g) + u(f(0)\cdot g) \;=\; u(f-f(0)\cdot g) + f(0)\cdot u(g) $$ is the linearity of $u$. As to evaluating the constant which involves the Euler-Mascheroni constant, I do not have an easy answer. But the literal integral can be manipulated in several ways, for example integrating by parts, to get something like your 'pf' functional.