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?
Consider the following figure:



Triangles $\triangle ABC$ and $\triangle ACD$ are right triangles, and $AB\cong BD$, so $\triangle ABD$ is isosceles with 

$$\angle BAD=\angle BDA={1\over2}(\pi-\angle ABD)={1\over2}\left(\pi-\left({\pi\over2}-\angle BAC \right)\right)={\pi\over4}+{1\over2}\angle BAC$$

where all angles are taken in a positive sense.  We also have

$$\angle CAD=\angle BAD-\angle BAC={\pi\over4}-{1\over2}\angle BAC$$

Finally, taking care with minus signs, we have

$$\angle CAD=-\tan^{-1}(x-\sqrt{1+x^2})\quad\text{and}\quad\angle BAC=\tan^{-1}x$$

which gives the desired identity

$$\tan^{-1}(x-\sqrt{1+x^2})={1\over2}\tan^{-1}x-{\pi\over4}$$

Remark (added later):  The picture and proof here assume $x\ge0$.  To cover the case $x\lt0$ requires either its own picture or an appeal to some kind of analytic continuity.
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 $x&gt;0$, $x-\sqrt{1+x^2}&lt;0$ since $1+x^2&gt;x^2$ implies $\sqrt{1+x^2}&gt;x$.

But it's also $&gt;-1$, since

$$(1+x)^2=x^2+1+2x&gt;x^2+1$$

And taking square root, $1+x&gt;\sqrt{1+x^2}$

Thus, for $x&gt;0$, $-1&lt;x-\sqrt{1+x^2}&lt;0$.

It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|&lt;1$:

That is, for $x&gt;0$,

$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$

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 $x&gt;0$,

$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$

And since $\arctan x+\arctan \frac1x=\pi/2$ for $x&gt;0$, you have

$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$



Now, we have to prove the same for $x&lt;0$.

Then, $x-\sqrt{1+x^2}&lt;-1$, so we can't apply our formula with arctangent.
However, its inverse is in $]-1,0[$, and for $x&lt;0$,

$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$

$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$

Then, still for $x&lt;0$, $x+\sqrt{1+x^2} \in ]0,1[$ 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 $x&lt;0$), 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 $x&gt;0$ and for $x&lt;0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.