Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$

The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests a promising method of finding a polynomial close to $\arctan$

*

*divide $1$ by $1+t^2$, to obtain a polynomial plus a remainder:

$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\tag{1}$$

How exactly do we perform this division of $1$ by $1+t^2$ to obtain $(1)$?

$$\frac{1}{1+t^2}= 1 - \frac{t^2}{1+t^2}$$ implies
$$\frac{1}{1+t^2}=1-t^2\left(1 - \frac{t^2}{1+t^2}\right)= 1 -t^2 + \frac{t^4}{1+t^2}$$ and continue in this manner.
