so i know the answer for $\int{\frac{x \, dx}{(x+b)^{2}}} \quad \textrm{is} \quad \frac{b}{x+b} + ln|x+b| $

But i tried integration by parts and obtained the following, Setting $ u= x, \, du=dx, \, dv = \frac{dx}{(x+b)^{2}}, \, v=\frac{-1}{x+b} $

$ \int{\frac{x \, dx}{(x+b)^{2}}}= \frac{-x}{x+b} + ln|x+b|$

Does it have anything to do with any possible singularities in the integrand?


1 Answer 1


Both results are true, they just differ by a constant :

$$ \frac{-x}{x+b} + \ln|x+b|=\frac{-x-b+b}{x+b}+\ln|x+b|=\color{red}{-1}+\frac{b}{x+b}+\ln|x+b|. $$

  • $\begingroup$ workaholic, how are you progressing with your linear algebra? $\endgroup$
    – abel
    Mar 7, 2015 at 2:48
  • $\begingroup$ This is not really a hint. $\endgroup$ Mar 7, 2015 at 3:08
  • $\begingroup$ And if it was a hint then it shouldn't be posted as an answer. $\endgroup$
    – user253751
    Mar 7, 2015 at 6:00
  • $\begingroup$ @FengyangWang I initially just wrote the first sentence, but after posting it I immediately edited my answer to include those equalities, forgetting to remove the hint. $\endgroup$
    – Workaholic
    Mar 11, 2015 at 19:39
  • $\begingroup$ Dear @abel $\!,$ Sorry, I didn't saw the message you sent me last time via e-mail, I'm sorry for that, I'll respond to it later. :-) (I was very busy these last days) $\endgroup$
    – Workaholic
    Mar 17, 2015 at 20:41

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