Power series expansion of $\text{arsinh}(x)$ at $0$

Question

How can one expand $\text{arsinh}(x)$ in a power series at $0$?

I know that $\text{arsinh}(x) = \ln(x+\sqrt{x^2+1})$

and that

$$\text{arsinh}(x)' = \frac{1}{\sinh'(\text{arsinh}(x))} = \frac{1}{\cosh(\text{arsinh}(x))}$$

$$=\frac{1}{\sqrt{1+\sinh^2(\text{arsinh}(x))}} = \frac{1}{\sqrt{1+x^2}}$$

$$\text{since } \cosh^2(x)-\sinh^2(x)=1 \Leftrightarrow \cosh^2(x) = 1+\sinh^2(x)$$

And if we define

$$f(x):=\text{arsinh}(x)-\ln(x+\sqrt{x^2+1})$$

Then

$$f'(x) = \frac{1}{\sqrt{1+x^2}} - \frac{1}{x+\sqrt{1+x^2}} \cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{1+x^2}} - \frac{1}{\sqrt{1+x^2}} = 0$$

But that just proved the identity. How do I get the power series expansion of $\text{arsinh}(x)$ at $0$?

On proof wiki I found this

Answer 1

Note that\begin{align}\frac1{\sqrt{1+x^2}}&=(1+x^2)^{-1/2}\\&=1-\frac12x^2+\frac{1\times3}{2!\times2^2}x^4-\frac{1\times3\times5}{3!\times2^3}x^6+\cdots\end{align}and that $\operatorname{arcsinh}(0)=0$. Therefore\begin{align}\operatorname{arcsinh}(x)&=\int_0^x\frac{\mathrm dt}{\sqrt{1+t^2}}\\&=\int_0^x\left(1-\frac12t^2+\frac{1\times3}{2!\times2^2}t^4-\frac{1\times3\times5}{3!\times2^3}t^6+\cdots\right)\,\mathrm dt\\&=x-\frac1{3\times2}x^3+\frac{1\times3}{5\times2!\times2^2}x^5-\frac{1\times3\times5}{7\times3!\times2^3}x^7+\cdots\end{align}