This entry in Wikipedia states the following theorem (http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Application_to_symbolic_integration)
Let $f$ and $g$ be nonzero polynomials over a field $K$. Write $g$ as a product of powers of pairwise coprime polynomiasls which have no multiple root in the algebraic closure of $K$:$$g=\prod_{i=1}^k p_i^{n_i}.$$There are (unique) polynomials $b$ and $c_{ij}$ with $\deg c_{ij} < \deg p_i$ such that $$\frac fg = b + \sum^k_{i=1}\sum^{n_i}_{j=2}\left(\frac {c_{ij}}{p_i^{j-1}}\right)' + \sum^k_{i=1} \frac {c_{1j}}{p_i}$$ where $'$ denotes the derivative, and the last summand is called the logarithmic part.
Indeed, $$\frac{c_{i1}}{p_i}=\sum_{\alpha_j:p_i(\alpha_j)=0}\frac{c_{i1}(\alpha_j)}{p'_i(\alpha_j)}\frac{1}{x-\alpha_j}.$$
My questions are two:
1) How does one get the last expression for the logarithmic part? According to the entry, this related to Lagrange interpolation, but I don't see how.
2) Under what conditions is one allowed to integrate the logarithmic part from the expansion I ask? I am confused, because if $K = \Bbb R$ and, say, $\frac fg = \frac 1{x^2 + 1}$, then one computes the integral to be $\arctan x$ and not $$\int \frac 1 {2i} \left(\frac {1} {x+i} - \frac 1 {x-i}\right) dx = \frac 1 {2i} (log (x+i) -log (x-i))$$