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The following integration code is identical, except for the use of integration variables x vs k and constants a vs b (all four possible combinations are included). Oddly, only the third one gives an output in terms of the ArcTan function, while all others give ArcTanh. Why?

This is obviously just a curiosity, since it doesn't really matter (both outputs are mathematically correct).

Assuming[a > 0, Integrate[(1/Sqrt[x^2 + 1])*(1/(x - a)), x]]
Assuming[a > 0, Integrate[(1/Sqrt[k^2 + 1])*(1/(k - a)), k]]
Assuming[b > 0, Integrate[(1/Sqrt[x^2 + 1])*(1/(x - b)), x]]
Assuming[b > 0, Integrate[(1/Sqrt[k^2 + 1])*(1/(k - b)), k]]

Output:

-((2*ArcTanh[(a - x + Sqrt[1 + x^2])/Sqrt[ 1 + a^2]])/Sqrt[ 1 + a^2])
-((2*ArcTanh[(a - k + Sqrt[1 + k^2])/Sqrt[ 1 + a^2]])/Sqrt[ 1 + a^2])
  (2*ArcTan [(b - x + Sqrt[1 + x^2])/Sqrt[-1 - b^2]])/Sqrt[-1 - b^2]
-((2*ArcTanh[(b - k + Sqrt[1 + k^2])/Sqrt[ 1 + b^2]])/Sqrt[ 1 + b^2])
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    $\begingroup$ Does this answer your question? Integration result depending on variable name $\endgroup$
    – creidhne
    Commented Jun 7 at 0:36
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    $\begingroup$ Check the tag letter-makes-difference and you'll find more related posts :) . As to your specific question, it's no longer the case in version 14.0: i.sstatic.net/pzN51hXf.png Which version are you in? $\endgroup$
    – xzczd ♦
    Commented Jun 7 at 0:50
  • $\begingroup$ @xzczd Interesting. I'm using V.13.2.1. $\endgroup$
    – WillG
    Commented Jun 7 at 1:28
  • $\begingroup$ @creidhne That's the same question, but the answer there is that it's fixed in V12, whereas I'm using V13. Apparently this case of the issue is fixed in V14, but still, my question is why it happens, not just what version it's fixed in. $\endgroup$
    – WillG
    Commented Jun 7 at 14:36

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