Question 1182692
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Many people,  when are asked about the integral of    {{{1/x}}},    mistakenly answer    ln(x) + Const.


This answer is  INCORRECT.


The  CORRECT  answer is    ln(|x|) + Const.




        And it works for  BOTH  positive and negative values of  x  (for all real values of  x  except of  0  (zero) ).




Correspondingly,   the integral of    {{{1/(x-a)}}}    is    ln(|x-a|) + Const1     (at x =/= a);


                            the integral of    {{{1/(x+b)}}}    is    ln(x+b|) + Const2      (at x =/= -b).