<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, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.
And for all $y\neq \pi/4+k\pi/2$,
$$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$
Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:
$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$
And since $2\arctan z \in ]-\pi/2,\pi/2[$, you can also apply the first identity to get
$$\arctan \tan (2\arctan z)=2\arctan z=\arctan \left(\frac{2z}{1-z^2}\right)$$
Bear in mind that this is proved only for $z\in]-1,1[$, and it actually is wrong outside this interval.
Now, for $x>0$, $x-\sqrt{1+x^2}<0$ since $1+x^2>x^2$ implies $\sqrt{1+x^2}>x$.
But it's also $>-1$, since
$$(1+x)^2=x^2+1+2x>x^2+1$$
And taking square root, $1+x>\sqrt{1+x^2}$
Thus, for $x>0$, $-1<x-\sqrt{1+x^2}<0$.
It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|<1$:
That is, for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$
And
$$\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}=\frac{2(x-\sqrt{1+x^2})}{-2x(x-\sqrt{1+x^2})}=-\frac{1}{x}$$
Thus, still for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$
And since $\arctan x+\arctan \frac1x=\pi/2$ for $x>0$, you have
$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$

Now, we have to prove the same for $x<0$.
Then, $x-\sqrt{1+x^2}<-1$, so we can't apply our formula with arctangent.
However, its inverse is in $]-1,0[$, and for $x<0$,
$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$
Then, still for $x<0$, $x+\sqrt{1+x^2} \in ]0,1[$ and
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{1-(x+\sqrt{1+x^2})^2}$$
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\arctan \frac1x$$
And $\arctan x+\arctan \frac1x=-\pi/2$ (remember that $x<0$), hence
$$2\arctan (x+\sqrt{1+x^2})=\arctan (x)+\pi/2$$
And finally
$$2\arctan (x-\sqrt{1+x^2})=2\arctan (x+\sqrt{1+x^2})-\pi=\arctan (x)-\pi/2$$
This formula is now proved for $x>0$ and for $x<0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.
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