<s>[INST] How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of 
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find 
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies 
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
 [/INST] First let's figure out what $C$ is. Letting $x=0$, we see that $\arctan(-1) = C$ (i.e. $C = -\frac{\pi}{4}$ or $C = \frac{3\pi}{4}$). Since inverse trigonometric functions are hard to work with, let's take $\tan$ of both sides. Doing so we get
$$\tan\left(\arctan(x-\sqrt{1+x^2})\right) = \tan\left(\frac{1}{2}\arctan(x)+C\right).$$
The left hand side is nothing more than $x-\sqrt{1+x^2}$. As for the right hand side, we can employ the tangent sum formula:
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}.$$
Thus
$$\tan\left(\frac{1}{2}\arctan(x)+C\right) = \frac{\tan\left(\frac{1}{2}\arctan(x)\right) + \tan(\arctan(-1))}{1-\tan\left(\frac{1}{2}\arctan(x)\right)\tan(\arctan(-1))}.$$
We have that $\tan\left(\frac{1}{2}\arctan(x)\right) = \frac{\sqrt{1+x^2}-1}{x}$ which can be verified by half angle formulas. If you put these pieces together and do some algebra, you will have your answer.
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