I have a problem understanding a step in the proof of the Sochozki formula, in the book: Vladimirov, Methods of the Theory of Generalized Functions. At page 20:
If $\varphi \in \mathcal{D}(\mathbb{R}),\; \varphi(x) = 0$ for $|x| > R$, then $$ \begin{align} \lim_{\epsilon \rightarrow 0^+} \int \frac{\varphi(x)}{x+i}\; dx &= \lim_{\epsilon \rightarrow 0^+} \int_{-R}^R \frac{x-i\epsilon}{x^2+\epsilon^2} \varphi(x)\; dx\\ &=\varphi(0) \lim_{\epsilon \rightarrow 0^+} \int_{-R}^R \frac{x-i\epsilon}{x^2+\epsilon^2} \; dx + \lim_{\epsilon \rightarrow 0^+} \int_{-R}^R \frac{x-i\epsilon}{x^2+\epsilon^2} [\varphi(x)-\varphi(0)]\; dx\\ &= -2i\varphi(0) \lim_{\epsilon \rightarrow 0^+} \arctan{\frac{R}{\epsilon}} +\int_{-R}^R \frac{\varphi(x)-\varphi(0)}{x}\;dx\\ \end{align} $$
I have some trouble figuring out who he did the last step and how he find the $\arctan$ term.
Separate the terms in the numerator to give two integrals
\begin{align} \int_{-R}^R \frac{x-i\epsilon}{x^2+\epsilon^2} \; dx &= \left[ \frac 12 \log(x^2+\epsilon^2) \right]_{-R}^R - \frac i\epsilon \int_{-R}^R \frac{1}{(x/\epsilon)^2+1} \; dx. \end{align}
That second term integrates to give the $\arctan$ term.