Primitive $r/(1+r^2)$ without abs()

Why should there not be an absolute value-sign instead of () when I find the primitive of $r/(1+r^2)$? Maybe it should only be there when I derive?

http://www.wolframalpha.com/input/?i=primitive+r%2F%281%2Br%5E2%29

• In principle there shoud be, but since $r^2+1> 0$, for all $r\in \mathbb R$, you can remove them. In any case, if I recall correctly, WA never includes $\text{abs}$ even when it's necessary. Apr 16, 2014 at 10:54
• Ok. Question solved then. Apr 16, 2014 at 11:14
• Feel free to answer it below, so the question doesn't come up as unanswered. Apr 16, 2014 at 11:15
• By default Mathematica and W|A work with complex functions. Now, $\log(z)$ is differentiable on the complex plane, except on the negative real axis where it has a branch cut. $\log(|z|)$, by contrast is nowhere differentiable as a complex plane. So, to assert that $\int(1/z)dz=\log(z)$ in this context is simply incorrect. Apr 16, 2014 at 15:10

In principle there shoud be, but since r2+1>0, for all r∈ℝ, you can remove them. In any case, if I recall correctly, WA never includes abs even when it's necessary. -- Git Gud Apr 16 at 10:54

• You should really learn to cite the people you copy stuff from. See also math.stackexchange.com/q/955806.
– Did
Oct 3, 2014 at 6:13
• There is nothing to cite here. It's a short question and a link to wolfram alpha. Oct 3, 2014 at 14:09
• Yes there is (in boldface or not): you posted as an answer a comment from user Git Gud, which you copied without attribution.
– Did
Oct 3, 2014 at 16:22