<s>[INST] Power Series Expansion for Real Inverse Hyperbolic Sine Context: independent self-study.
In Murray R. Spiegel's Mathematical Handbook of Formulas and Tables (1968) we have in Chapter $20$ Taylor Series, section "Series for Hyperbolic Functions":
$20.39 \quad \sinh^{-1} x = \begin{cases}x - \dfrac {x^3} {2 \cdot 3} + \dfrac {1 \cdot 3 x^5} {2 \cdot 4 \cdot 5} - \dfrac {1 \cdot 3 \cdot 5 x^7} {2 \cdot 4 \cdot 6 \cdot 7} + \cdots& |x| < 1 \\
\pm \left({\ln \left|2 x\right| + \dfrac 1 {2 \cdot 2 x^2} - \dfrac {1 \cdot 3} {2 \cdot 4 \cdot 4 x^4} + \dfrac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 \cdot 6 x^6} - \cdots}\right) & \text{$+$ if $x \ge 1$, $-$ if $x \le -1$} \end{cases}$
The first part is straighforward enough, expand $\dfrac 1 {\sqrt {x^2 + 1}}$ and integrate term by term, justified by appropriate theorems in real analysis.
But getting to the second part, where $|x| \ge 1$, is more of a challenge.
My initial thought is to add $\sinh^{-1} x$ to $\sinh^{-1} \left(\dfrac 1 x \right)$ because the bit in the brackets looks promisingly like the expansion of the reciprocal of that.
Using the logarithmic presentation of $\sinh^{-1} x$ you get $\ln (x + \sqrt {x^2 + 1} )$, and for $\sinh^{-1} \left(\dfrac 1 x \right)$  you have $\ln \left(\dfrac 1 x + \dfrac {\sqrt {x^2 + 1} } x \right)$. Adding those terms is logarithm of product so multiplying $\ln \left( (x + \sqrt {x^2 + 1} ) \left(\dfrac 1 x + \dfrac {\sqrt {x^2 + 1} } x \right) \right)$. Multiplying that out should give something simple, but it doesn't go down simpler than $\ln \left(1 + \dfrac 1 x \sqrt {x^2 + 1} + \sqrt {x^2 + 1} + \dfrac {x^2 + 1} x\right)$, where I wanted to find $\ln (2 x)$ or something.
Any hints as to what manipulation I need to get that expression for $\sinh^{-1} x$ for $|x| \ge 1$?
 [/INST] Just move the $\ln |2x|$ to the LHS. We have (if $x>0$)
$$
\sinh^{-1} (1/x) + \ln (2x) = \ln 2 + \ln(1+\sqrt{1+x^2}). 
$$
It remains to expand $f(x)=\ln(1+\sqrt{1+x^2})$ around $x=0.$ Let's leave this as an exercise :)
Hint: consider $g(x) = \ln(1+\sqrt{1+x}).$ Take derivate $g'(x).$ Clear square roots in the denominator. Is it much simpler now?
The case $x<0$ is similar.
</s>