Find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$ I need to find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$ and present the series as a sum of members with the same formula. Since we haven't studied integrals yet, I'd prefer suggestions without them. For the Maclaurin series, we studied the formula $f(x)=f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^{2}}{2!}+...+\frac{f^{(n)}(0)x^{n}}{n!}+o(x^{n})$. From this formula, we derived $\ln{(1+x)}=\sum_{k=1}^{n}{\frac{(-1)^{k-1}x^{k}}{k}}+o(x^{n})$. So, trying to use one of these, I thought of two approaches:

*

*$\ln{(x(1+\frac{\sqrt{x^2+1}}{x}))}=\ln{x}+\ln{(1+\frac{\sqrt{x^2+1}}{x})}$ but then that root fraction is not suitable for the $\ln$ formula

*the derivative of $\ln{(x+\sqrt{x^2+1})}$ is $(1+x^{2})^{-\frac{1}{2}}$. But then I will have to find the n-th derivative for the basic formula, which I wouldn't be able to. How should I procede?

 A: Not a full answer but a few steps you will encounter on the way to one. There is a pattern if you keep differentiating, and keep rearranging to get the following landmarks, which will keep it all manageable.
Show that, (as you have already done in your method 2)
$$\sqrt{1+x^2}f'(x)=1$$
then (using the product rule on the line above) that,
$$(1+x^2)f''(x)+xf'(x)=0$$
then (using the product rule on the line above),
$$(1+x^2)f'''(x)+3xf''(x)+f'(x)=0$$
and so on, as required...
You can then extract the values of $f(0), f'(0),f''(0), f'''(0),...$ and use them in the formula that you tell me you have studied.
This method, although long, avoids integration, as you requested.
A: You don't want to use integration, but I don't think you can avoid it entirely.  First note that
$$f(x) = \ln(x+\sqrt{x^2+1}) = \mbox{arcsinh } x, $$
and as you note
$$f'(x) = (1+x^2)^{-1/2}.$$
The generalized binomial theorem
https://en.wikipedia.org/wiki/Binomial_theorem#Newton's_generalized_binomial_theorem
gives you
$$f'(x) = \sum_{n=0}^{\infty} \frac{ \left(-\frac{1}{2}\right)_n}{n!}x^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n(2n)!}{2^{2n}(n!)^2}x^{2n}.$$
Since you know how to differentiate polynomials, then it's just barely "integration" to un-differentiate this series and get your answer.  Realize that any constant could be added to your answer and still differentiate to the above.  So you can plug in $x=0$ at the end and work out that constant.  (It'll be $0$.)
