<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] By virtue of solving for $x$, the original equation $$y = \tan^{-1} (x - \sqrt{1+x^2})$$ implies $$x = \frac{1}{2}(\tan y - \cot y) = \frac{\sin^2 y - \cos^2 y}{2 \sin y \cos y} = -\cot 2y,$$ hence $$y = \frac{1}{2}\cot^{-1} (-x) = -\frac{1}{2} \cot^{-1} x.$$  Now recalling that $$\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2},$$ we readily obtain $$y = \frac{1}{2} \tan^{-1} x - \frac{\pi}{4}.$$
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