why does integration by parts give a different answer

Question

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?

Workaholic · Answer 1 · 2015-03-11T19:37:11.340

14

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

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

edited Mar 11 '15 at 19:37

answered Mar 06 '15 at 22:40

Workaholic

workaholic, how are you progressing with your linear algebra? – abel Mar 07 '15 at 02:48
This is not really a hint. – Fengyang Wang Mar 07 '15 at 03:08
And if it was a hint then it shouldn't be posted as an answer. – user253751 Mar 07 '15 at 06:00
@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. – Workaholic Mar 11 '15 at 19:39
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) – Workaholic Mar 17 '15 at 20:41

1 Answers1