Trigonometric proof using derivatives

[Pole]

Homework Statement

Prove that (for every x smaller than -1)
\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}}+arccotx=\pi

Homework Equations

The Attempt at a Solution

So i split the formula into two parts:
\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}} and \displaystyle \pi-arccotx
Calculated their derivatives separately
And they are both equal to \displaystyle \frac{1}{x^{2}+1}

And I'm wondering whether it's the end of the task or should I prove some other things, especially something connected to the main assumption of x<-1 or any other.

Thanks in advance!

 

[micromass]

If f^\prime=g^\prime, then this doesn't mean f=g necessarily. But it does mean that f=g+C with C a real constant/

Note that the above only holds if the domain of f and g is an interval.

 

[Pole]

If f^\prime=g^\prime, then this doesn't mean f=g necessarily. But it does mean that f=g+C with C a real constant/

Note that the above only holds if the domain of f and g is an interval.

Thank You, now I know that it isn't sufficient.
But I can't really come up with a complete solution.
What should I add ? So that the task would be marked well
How to calculate the value of that constant?