# Series expansion of $\left(1 + x^2\right)^{-1/2}$

I have the following function: $$f(x) = arcsinh(x)$$ which i want to express as a series of powers. To do that I thought: $$\frac{df}{dx} = \frac{1}{\sqrt{1 - x^2}}$$ so $$f = \int\frac{1}{\sqrt{1 - x^2}}dx= \int\left(1 - x^2\right)^{-1/2}dx$$ and now i will expand that using the binomial expansion $(1 + x)^a$. I tried several ways to make that expansion but I couls not get a compact expression. After some search I saw that: $$\left(1 - x\right)^{-1/2} = \sum_{k=0}^{\infty}\frac{(-1)^k(2k)!}{2^{2k}(k!)^2}x^k$$ But I dont understand how this expression was obtained.Any other ideas for expressing my function as a series of powers or a demonsstration as to how this expression is obtined?

$$(1+x^2)^{-1/ 2} = \sum_{n=0}^{\infty} \binom{-1/ 2}{n} x^{2n}$$
then (fixing your sign mistake), where the power series about $0$ for $\arcsin(x)$ is defined,
$$\arcsin(x) = \int \frac{dx}{(1-x^2)^{1/ 2}} = \int\sum_{n=0}^{\infty} \binom{-1/ 2}{n} (-x^2)^{n} = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \binom{-1/ 2}{n}x^{2n+1}$$ and you can do the rest...