integration - Why does $dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8dots$?

I apologize if this is a duplicate.

I was taught how to prove that $\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}\dots$, and one of the steps was to write the equality:
$$\int \dfrac{1}{1+x^2} \ dx = \int \sum\limits_{n=0}^\infty x^{2n}\cdot(-1)^n \ dx = \int 1-x^2+x^4-x^6+x^8\dots \ dx$$
Why does $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots$? I have no idea on how to proceed with this. Could someone please point me in the right direction? Thanks in advance.

Answer

First, use this basic fact from geometric series:
$$
\frac1{1-x}=\sum_{n=0}^\infty x^n.
$$
Make the substitution $-x^2$ for $x$ to obtain
$$
\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}.
$$

EDIT: to elucidate the first bit, suppose have the infinite series

$$
a+r\cdot a+r^2\cdot a+\ldots
$$
where $|r|<1$. Let $L$ be this sum, supposing it exists; $|r|<1$ is actually a necessary and sufficient condition. Then $L-a=r\cdot L$ by construction, so we have
$$
a=(1-r)L\implies L=\frac a{1-r}.
$$
In our case, we have $a=1$, $r=x$.