Integrate $1/x$ in the sense of distribution, with non-symmetric excluded neighborhood of zero 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?  
 A: 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)
