<s>[INST] Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \dfrac{\pi}{4} $$
My question is: why is this constant and what procedure is done to discover it.
 [/INST] It's even nicer if you use simple trigonometry. 
Consider the following Figure, for positive $x$ (for negative $x$ see edit, thanks to Steven for pointing this out in comment).
$\hskip1.in$ 
The triangle $OAB$ is right-angled, and such that $\overline{OA} = 1$, and $\overline{AB} = x$. We have
\begin{equation}\angle BOA = 2\alpha = \arctan x\tag{1}\label{eq:0}\end{equation}
and 
$$\overline{OB} = \sqrt{x^2+1}.$$
Extend $BA$ to a segment $BC$ such that 
$$ BC \cong OB.$$
Then
$$\overline{AC} = \sqrt{1+x^2}-x,$$
and so
\begin{equation}\beta = \angle AOC = \arctan\left(\sqrt{1+x^2}-x\right).\tag{2}\label{eq:0p}\end{equation}
By considering the right-angled triangle $OAB$, you
obtain
\begin{equation}\angle OBA =  \frac{\pi}{2}-2\alpha,\tag{3}\label{eq:1}\end{equation}
and, by considering the isosceles triangle $OBC$, you get
\begin{equation}\angle OBA = \pi - 2(2\alpha+\beta).\tag{4}\label{eq:2}\end{equation}
Equating \eqref{eq:1} and \eqref{eq:2} yields
$$\alpha = -\beta + \frac{\pi}{4}.$$
Plugging in \eqref{eq:0} and \eqref{eq:0p} and using the odd symmetry of the $\arctan(\cdot )$ function leads to
$$ \frac{1}{2}\cdot \arctan x = \arctan\left(x-\sqrt{1+x^2}\right) + \frac{\pi}{4},$$
the desired result. $\blacksquare$
EDIT
The same result applies if $x<0$. Use then the Figure below.
$\hskip1.in$ 
Now $\overline{AB} = -x$, $\overline{OB} = \overline{BC} = \sqrt{1+x^2}$, and $\overline{CA} = \sqrt{1+x^2}-x$. Then, again using right-angled triangle $OAC$ and isosceles triangle $OBC$, we get
$$\angle OCA = \frac{\pi}{2}-\beta = \beta - 2\alpha,$$
where $2\alpha = \angle BOA$, and $\beta = \angle COA$. The  equality of OP follows.
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