Is there a trick to finding the Maclaurin series for $f(x)=\ln(x+\sqrt{1+x^2})$ fast? Vaguely, I recall this being some sort of inverse hyperbolic function, but I'm not sure about which one, and what its derivatives are. This is a past exam question and I would like to know, should something similar appear again, if there is a quick method of finding this series without using inverse hyperbolic functions. The derivatives of this look absolutely painful to calculate, although easy at $x=0$, but I'm not sure if I could easily see a pattern for the $n$-th derivative at $0$.
[Math] Maclaurin Series for $\ln(x+\sqrt{1+x^2})$
calculuspower series
Best Answer
The easiest method: note that $$ \frac{d}{dx}\left[\ln(x+\sqrt{1+x^2})\right]=\frac{1}{\sqrt{1+x^2}}=(1+x^2)^{-1/2}. $$ This last can be expressed through a binomial series; you can then integrate term by term, and solve for the constant of integration, to get the series you want.