About the relation $\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2$ It is well known that 
$$\int\frac{1}{1+x^2}\,\mathrm{d}x=\tan^{-1}x+C \tag{1}$$ 
However, I integrated this differently and got an unusual result. 
Suppose we make the substitution $x=\sinh\theta$ and $\mathrm{d}x=\cosh\theta\,\mathrm{d}\theta$ so the integral becomes $$\int\frac{\cosh\theta}{\cosh^2\theta}\,\mathrm{d}\theta=\int\frac{1}{\cosh\theta}\,\mathrm{d}\theta \tag{2}$$
By the definition of $\cosh\theta$, we can rewrite this as $$\int\frac{2e^\theta}{e^{2\theta}+1}\,\mathrm{d}\theta=2\tan^{-1}e^\theta+C \tag{3}$$
Using the fact that $e^\theta=\cosh\theta+\sinh\theta$, we get $e^\theta=x+\sqrt{1+x^2}$, so the answer is then $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C \tag{4}$$ 
Equating $(4)$ with $(1)$, we have $$2\tan^{-1}\left(x+\sqrt{1+x^2}\right)+C=\tan^{-1}x \tag{5}$$
Plugging in $x=0$, we find $C=-\frac\pi2$. We now have the following strange relationship

$$\tan^{-1} x= 2\tan^{-1}\left(x+\sqrt{1+x^2}\right)-\frac\pi2 \tag{$\star$}$$
  This leads me to wonder: Why is this true geometrically, and does this relationship extend into the complex plane?

 A: The geometric interpretation could be found in the figure below.
We have $\alpha = \frac{\pi}{2} - \arctan x , ~\beta = \frac{\pi}{2} - \arctan \left(x + \sqrt{1 + x^2}\right)$, and $\alpha = 2 \beta$.
Combining these three equations would lead to your results. 

For a formal proof, let $\theta = \arctan \left(x + \sqrt{1 + x^2}\right)$. Since
$$ x + \sqrt{1 + x^2} \geq \sqrt{1 + x^2} - |x| > 0,$$
we have $\theta \in (0, \frac{\pi}{2})$. 
Also, note that
$$
\begin{align}
\tan\left(2\theta - \frac{\pi}{2}\right)&= -\frac{1}{\tan(2\theta)}\\
&= -\frac{1 - \tan^2 \theta}{2 \tan \theta}\\
&= -\frac{1 - \left(x + \sqrt{1 + x^2}\right)^2}{2 \left(x + \sqrt{1 + x^2}\right)}\\
&= x.
\end{align}
$$
As $\left(2\theta - \frac{\pi}{2}\right) \in (-\frac{\pi}{2}, \frac{\pi}{2})$, i.e., $\left(2\theta - \frac{\pi}{2}\right)$ lies in the range of $\arctan(\cdot)$, we know that
$$ \arctan x = 2\theta - \frac{\pi}{2} = 2\arctan \left(x + \sqrt{1 + x^2}\right) - \frac{\pi}{2}. $$
A: Since
$\arctan(x)+\arctan(y)
=\arctan(\frac{x+y}{1-xy})
$
with $k\pi$
thrown in as needed,
$\begin{array}\\
\arctan x- 2\arctan\left(x+\sqrt{1+x^2}\right)
&=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(x+\sqrt{1+x^2})^2}\right)\\
&=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(x^2+2x\sqrt{1+x^2}+1+x^2)}\right)\\
&=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{1-(2x^2+2x\sqrt{1+x^2}+1)}\right)\\
&=\arctan x- \arctan\left(\frac{2x+2\sqrt{1+x^2}}{-(2x^2+2x\sqrt{1+x^2})}\right)\\
&=\arctan x- \arctan\left(\frac{x+\sqrt{1+x^2}}{-x(x+\sqrt{1+x^2})}\right)\\
&=\arctan x- \arctan\left(\frac{1}{-x}\right)\\
&=\arctan x+ \arctan\left(\frac{1}{x}\right)\\
&=\arctan\left(\frac{x+\frac1{x}}{1-x\frac1{x}}\right)\\
&=\arctan\left(\frac{x+\frac1{x}}{0}\right)\\
&=\frac{\pi}{2}\\
\end{array}
$
A: We need to be very careful with principal values ,
Let arccot$(x)=2y\implies x=\cot2y$ and $0<2y<\pi$
$x+\sqrt{1+x^2}=\dfrac{\cos2y+1}{\sin2y}=\cot y$ as $y\ne\dfrac\pi2$
$\implies2\tan^{-1}(x+\sqrt{1+x^2})=2\left(\dfrac\pi2-\cot^{-1}(x+\sqrt{1+x^2})\right)=\pi-2y$
and $\tan^{-1}x=\dfrac\pi2-\cot^{-1}x=\dfrac\pi2-2y$
