Arctan is the inverse operation of tan, in the sense that $\arctan \tan \theta = \theta$.
To prove this, we note that $y = \arctan x$ implies that $\tan y = x$. So that $\rm \frac{d}{dx} \arctan x = \dfrac{1}{\frac{d \tan y}{dy}} = \frac{1}{\sec ^2 y} = \frac{1}{1 + \tan ^2 y} = \frac{1}{1 + x^2}$
And that is exactly what your book says.
---EDIT--- Listening to Hardy's suggestion, I add the following. Suppose we have two functions, $f$ and $g$ s.t. $f(g(x)) = g(f(x)) = x$, i.e. they are inverses of each other. Then by the chain rule, differentiating $f(g(x)) = x$, we get $\rm\frac{df}{dg} \frac{dg}{dx} = 1$, and this implies that $\rm \frac{dg}{dx} = \dfrac{1}{\frac{df}{dg}}$. I used this implicitly in my transition from the first to the second parts of the equation above.