Why, although these functions have the same derivative, do they not differ by a constant?
Solution 1:
The problem is that $\arctan \frac{1+x}{1-x}$ isn't defined at $x = 1$, and in particular isn't differentiable there. In fact, we have $$ \arctan \frac{1+x}{1-x} - \arctan x = \begin{cases} \frac{\pi}{4} & x < 1, \\ -\frac{3\pi}{4} & x > 1. \end{cases} $$ So the difference is piecewise constant.
Solution 2:
Hint: $$\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x\tan y}$$
What happens when $\tan y=1$?
Solution 3:
Piecewise, they do differ by a constant. Just with a discontinuity at x=1.
And we explicitly define arctan(t) to have that discontinuity at t=+/-inf