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I'm sure this is quite a simple question, but am I right that

$$ \int\mathrm dx\frac{C}{x-C} = C \int\mathrm dx\frac{1}{x-C} = C\,\mathrm{log}\left(|x-C|\right) + \mathrm{const.} $$

If yes, then why does Anthony Zee get $p-q=2r+2r_S\,\,\mathrm{log}\frac{|r-r_S|}{r_S}$ on page 424 of his Einstein Gravity in a Nutshell? Why the $r_S$ term in the denominator inside of the logarithm?

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It isn't listed in the errata list, and the denominator is later used to cancel an $r_S$, so I'm pretty sure it's not a typo.

Andy Miles
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Both you and Zee are right: remember that this is indefinite integration, so you have no way to choose the integration constant; any such constant is equally good. Now, Zee's result can be rewritten as $2 r_s \log |r - r_s| - 2 r_s \log r_s$, and $2 r_s \log r_s$ is just a constant (with respect to $r$), so you may absorb it into your own $\text{const}$.

Alex M.
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