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**Prove It** Well the way I would think about it is to note that

$\displaystyle \frac{d}{dx}(\arctan{x}) = \frac{1}{1 + x^2}$.

You should know that for a geometric series $\displaystyle \sum_{n = 0}^{\infty}ar^n$, its infinite sum is $\displaystyle \frac{a}{1 - r}$ provided $\displaystyle |r| < 1$.

I think you can see that if you write $\displaystyle \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \frac{a}{1 - r}$ then you can see $\displaystyle a = 1, r = -x^2$.

So $\displaystyle \frac{1}{1 + x^2}$ is the closed form of $\displaystyle \sum_{n = 0}^{\infty}1(-x^2)^n = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$. This is provided of course that $\displaystyle |r| < 1$.

So $\displaystyle |-x^2| < 1$

$\displaystyle |x|^2 <1$

$\displaystyle |x| < 1$.

So we have

$\displaystyle \frac{1}{1 + x^2} = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $\displaystyle |x| < 1$

$\displaystyle \frac{d}{dx}(\arctan{x}) = \sum_{n = 0}^{\infty}(-1)^nx^{2n}$ for $\displaystyle |x| < 1$

$\displaystyle \arctan{x} +C= \int{\sum_{n = 0}^{\infty}(-1)^nx^{2n}\,dx}$ for $\displaystyle |x| < 1$ (*)

$\displaystyle \arctan{x} = \sum_{n = 0}^{\infty}\int{(-1)^nx^{2n}\,dx}$ for $\displaystyle |x|<1$

$\displaystyle \arctan{x} = \sum_{n = 0}^{\infty}\frac{(-1)^nx^{2n + 1}}{2n + 1}$ for $\displaystyle |x| < 1$.