This is a little bit rough and ready, but could hopefully be tidied up. First, as @paul points out, we need to agree on what we mean by a distribution. I'll be a little less generous and say that it is an object that acts on test functions to give a number. This action takes the form $$<\Delta,u> = \int_{-\infty}^{+\infty} \Delta(x)u(x) \,dx,$$
where $\Delta$ is the distribution and $u$ is a test function - that is, a smooth function ($C^\infty$) which vanishes outside some compact set. (Hence, there is a number $K>0$ such that $u$ and all its derivatives equal zero for $|x|>K$.) We also need to agree that if we can define functions $\Delta_\epsilon, \epsilon>0$ with the property that
$$\lim_{\epsilon\to 0}<\Delta_\epsilon,u> \quad \hbox{exists for all test functions }\,u,$$
then we say that $\Delta=\lim_{\epsilon\to 0}\Delta_\epsilon$ is a distribution with $$<\Delta,u>=\lim_{\epsilon\to 0}<\Delta_\epsilon,u>.$$ (These are standard definitions.)
So taking up your principal value argument, define, for $\epsilon>0$, $$\delta_\epsilon(x) = \int_{-\infty}^{x-\epsilon}\frac{dt}{t(t-x)}+\int_{x+\epsilon}^{+\infty}\frac{dt}{t(t-x)}.$$ We can calculate that
$$\delta_\epsilon(x) = \frac{1}{x}\log\left|\frac{x+\epsilon}{x-\epsilon}\right|.$$ [Oops. This doesn't deal correctly with $x=0$, so the rest of this answer is pretty much useless.]
(This - $\delta_\epsilon$ - is locally integrable on $\mathbb{R}$, which is the condition required of the 1-parameter ($\epsilon$) family of functions $\Delta_\epsilon$ above. That is, $\Delta_\epsilon$ is not required to be a classical function, which is useful because $\delta_\epsilon$ blows up at $x=\epsilon$. However the blow-up is very mild.)
So for the programme above to work, we have to check that (i) $<\delta_\epsilon,u>$ exists for all test functions $u$; (ii) the limit $\lim_{\epsilon\to 0}<\delta_\epsilon,u>$ exists. If in addition we can show that this limit evaluates to $cu(0)$ for some constant $c$ and all test functions $u$, we can then legitimately say that $\delta=\lim_{\epsilon\to 0}\delta_\epsilon$ is indeed equal to $c\delta_{Dirac}=c\times$ the Dirac delta.
So pick a test function $u$. Since $u$ is smooth everywhere, we can apply Taylor's theorem and write $$u(x)=u(0)+u'(0)x+\frac{u''(\xi_x)}{2}x^2,$$
where $\xi_x$ depends continuously on $x$. Then
$$<\delta_\epsilon,u> = I_1 + I_2 + I_3,$$
where
\begin{eqnarray*}I_1&=&u(0)\int_{-\infty}^{\infty}\delta_\epsilon(x)dx,\\
I_2&=&u'(0)\int_{-\infty}^{\infty}x\delta_\epsilon(x)dx,\\
I_3&=&\int_{-\infty}^{\infty}\frac{u''(\xi_x)}{2}x^2\delta_\epsilon(x)dx.\\
\end{eqnarray*}
Notice that $\delta_\epsilon$ is even, so $I_2=0$ and $I_1$ and $I_3$ can be written as integrals over the positive half-line. Now recall that $u$ and its derivatives vanish for all $|x|>K$. Let $M$ be the maximum of the second derivative of $u$ on $[-K,K]$. Then
$$|I_3|\leq M \int_0^K x^2\delta_\epsilon(x)dx =M\int_0^K x\log\left|\frac{x+\epsilon}{x-\epsilon}\right|dx.$$
We can evaluate this explicitly (bit of work...) and show that (i) it exists for $\epsilon>0$ and (ii) vanishes in the limit $\epsilon\to 0$.
It remains to deal with the integral $I_1$. We have
\begin{eqnarray*}
\int_0^\infty\frac{1}{x}\log\left|\frac{x+\epsilon}{x-\epsilon}\right| &=&
\int_0^\epsilon\frac{1}{x}\log\left(\frac{\epsilon+x}{\epsilon-x}\right) +
\int_\epsilon^\infty \frac{1}{x}\log\left(\frac{x+\epsilon}{x-\epsilon}\right)\\
&=& 2\int_0^1\frac{1}{t}\log\left(\frac{1+t}{1-t}\right)dt,\end{eqnarray*}
by using $x=\epsilon t$ in the first integral and $x=\epsilon/t$ in the second. This integral is finite, with value $c\simeq 4.93$ (probably something much more elegant, but this is what Wolfram Alpha gives).
So for any test function $u$, it seems that
$$\lim_{\epsilon\to0} <\delta_\epsilon,u> = c u(0),$$
and so your integral is $c\delta_{Dirac}$.
