Mathematica Collecting terms in mathematica

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To collect terms in Mathematica based on specific poles in the denominator, the user needs to utilize the `Select` function to filter terms containing those poles. The example provided demonstrates how to extract terms with a specific pole, such as (x-1), from a list of expressions. The user clarified that their expression is multi-dimensional and involves multiple products in the denominators, complicating the collection process. They found success in isolating terms by referencing the structure of their list. Ultimately, the discussion emphasizes the importance of understanding list structures and using appropriate functions in Mathematica for term collection.
dimwatt
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I have an integrand with a handful of terms, and some of them have poles in the denominator of the form (x+c) (but not all). There are three poles in total, and I want to collect all the terms according to each pole individually (eg all the terms with (x-1) in the denominator, (x-5), etc.)

How could I do this? I tried using collect just to see if mathematica would make any rearrangements at all, but I'm not having much luck with it.
 
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Without an example of what you have or what you have tried it is difficult to guess what to say.

Is this anything like what you want to do?

In[1]:= Collect[4/(x-g) + (a+b^2)/(x-f) + 2 q/(x-f) - 5/(x-f) - c/(x-g), {1/(x-f), 1/(x-g)}]

Out[1]= (-5 + a + b^2 + 2 q)/(-f + x) + (4 - c)/(-g + x)
 
Hi Bill!

Sorry I should have been much more specific. Each term in my expression involves several products in their denominators, and of those terms, some of them have poles like (x-1) and (y-1) and sometimes both, like N/(x+1)(y+1)(x-1)... for example (this is a multi-dimensional integral). I'm a bit new to mathematica so my terminology or descriptions might be a bit weird.. but basically I have the expression in list form, and I found that, e.g.,

Select[expr, MemberQ[#[[2]], (-1 + x)] &]

(the [[2]] is just a detail regarding the arrangement of the list) picks out the terms I was looking for. Thanks!
 

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