Show that for all real $x$:
$$\arctan(x+1) - \arctan(x) = \arctan(\frac{1}{x^2 + x +1})$$
This is what I got so far: $$\arctan(x+1/x) = \arctan(1/x^2 + x +1) \iff \\ x+1/x = 1/(x+1)^2 - x \iff \\ (x+1)^3-x^2-2x = 0$$
But then I get stuck and feel like there is something I'm missing or there is some smarter way of doing it.
