I'm kind of confused with this question because of the boundary terms. for a test function $\phi$ define: $ u(\phi) = \lim_{\epsilon \rightarrow 0+} \int_{x\notin(-3\epsilon,5\epsilon)} \phi(x) x^{-1} dx $
Find a locally integrable function $v$ such that $v'=u$ in the sense of distributions.
How can I deal with boundary terms to find the correct distribution?
The goal is to find a locally integrable $v$ such that $$\lim_{\epsilon \to 0+} \int_{x\notin(-3\epsilon,5\epsilon)} \phi(x) x^{-1} dx = -\int_{\mathbb{R}} \phi'(x)v(x)\,dx$$ As yousuf soliman suggested, use integration by parts, where the boundary term at $\infty$ does not appear because $\phi$ has compact support: $$ \int_{5\epsilon}^\infty \phi(x) x^{-1} dx = -\phi(5\epsilon) \log(5\epsilon) - \int_{5\epsilon}^\infty \phi'(x) \log x\, dx $$ and $$ \int_{-\infty}^{-3\epsilon} \phi(x) x^{-1} dx = \phi(-3\epsilon) \log(3\epsilon) - \int_{-\infty}^{-3\epsilon} \phi'(x) \log x\, dx $$ Add and let $\epsilon\to 0$. The integrals are okay because $\log |x|$ is integrable near $0$, but the boundary term requires some care. Subtracting off $\phi(0)$ helps here:
$$ \phi(-3\epsilon) \log(3\epsilon) -\phi(5\epsilon) \log(5\epsilon) = (\phi(-3\epsilon)-\phi(0)) \log(3\epsilon) - (\phi(5\epsilon) -\phi(0) \log(5\epsilon) + \phi(0) \log (3/5) $$ The first two terms are $O(\epsilon \log \epsilon)$, so they tend to $0$. The limit is $ \phi(0) \log (3/5) $. Summary of what we have so far:
$$u = \log(3/5) \delta_0 + (\log |x|)'$$ It remains to recall that the Dirac function $\delta_0$ is the distributional derivative of the Heaviside function $H$. $$v(x) = \log (3/5) H(x) + \log |x|$$ (up to an additive constant, of course)