# why does integration by parts give a different answer

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?

$$\frac{-x}{x+b} + \ln|x+b|=\frac{-x-b+b}{x+b}+\ln|x+b|=\color{red}{-1}+\frac{b}{x+b}+\ln|x+b|.$$
• 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) Mar 17, 2015 at 20:41