TITLE: Prove that $\arctan\left(\frac{2x}{1-x^2}\right)=2\arctan{x}$ for all $|x|<1$, directly from the integral definition of $\arctan$
QUESTION [5 upvotes]: I would like to show that for $A(x) = \int_{0}^{x}\frac{1}{1+t^2}dt$, we have $A\left(\frac{2x}{1-x^2}\right)=2A(x)$, for all $|x|<1$.
My idea is to start with either $2\int_0^x\frac{1}{1+t^2}dt$ or $\int_0^{2x/(1-x^2)}\frac{1}{1+t^2}dt$, and try to transform one into the other by change of variables. (It would make more sense for the moment if we did not do any trigonometric substitutions, since we are defining the trig functions via this integral.)
One of the several things I've tried is to use $A(x)+A(1/x)=\pi/2$, and write $\int_0^{2x/(1-x^2)}\frac{1}{1+t^2}dt=\pi/2 - \int_0^{(1-x^2)/2x}\frac{1}{1+t^2}dt=\pi/2 - \int_0^{1/2x}\frac{1}{1+t^2}dt+\int_0^{x/2}\frac{1}{1+t^2}dt$, but this doesn't seem to be getting me anywhere.
Any ideas?

REPLY [7 votes]: Hint: Let $f(x)=A\left(\frac{2x}{1-x^2}\right)$ and let $g(x)=2A(x)$, both defined by integrals precisely as in the OP. 
Use the Fundamental Theorem of Calculus to show that these functions have the same derivative. So they differ by a constant. Then all you will need to do is to show that they agree at say $x=0$, so the constant is $0$.