Proving $\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^2}}$, starting from the integral representation of $\arctan$ Starting from the integral representation of the arctan function, $$\arctan x = \int_0^x \frac{dt}{1+t^2},$$how does one prove the (half-angle) identity? $$\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^2}}$$
I am interested in a clever change of variables that should do the trick. However, I am unable to find it.

If you differentiate$$2\arctan\left(\frac t{1+\sqrt{1+t^2}}\right),$$then you get $\frac1{1+t^2}$. So\begin{align}2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)&=2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)-2\arctan\left(\frac0{1+\sqrt{1+0^2}}\right)\\&=\int_0^x\frac{\mathrm dt}{1+t^2}\\&=\arctan(x).\end{align}
