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?

Let me rewrite the proof a bit differently.
We are required to prove (RTP) that $$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\frac{\pi}{2}$$
Proof:
Plan: We will use the formula F first, followed by the identity I.
Let us call the following identity I:
$\tan^{-1}(x) + \tan^{-1} (1/x) =$
\begin{cases}
\hfill \pi/2 & \text{if $x\geq 0$}\\
      -\pi/2 & \text{if $x<0$}
\end{cases}
Next, call the following formula F:
$2 \tan^{-1} (x) =$
\begin{cases}
\hfill \tan^{-1} \frac{2x}{1-x^2} & \text{if $|x|\neq 1$}\\ 
       \frac{\pi}{2} & \text{if $x = 1$}\\
       -\frac{\pi}{2} & \text{if $x = -1$}
\end{cases}
Also note that $x - \sqrt{1+x^2} < 0$, 
therefore $|x - \sqrt{1+x^2}| > 1$ is equivalent to  $x - \sqrt{1+x^2} < -1 \implies x < 0$
Done with the ground work!
If $y < 0$, 
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (second case)]
If $y > 0$, 
LHS of RTP =
$$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\tan^{-1}\frac{1}{y} - \tan^{-1}(y) = -\frac{\pi}{2}$$
[we used formula F (first case) first, followed by identity I (first case)]
The case $y = 0$ is obvious!
