$$\ln (x+\sqrt{x^2+1})$$
I don't know how to proceed in a way other than finding the successive derivatives of this function.
First, I was trying to make the substitution $1 + t = x+\sqrt{x^2+1}$ but after writing the Maclaurin Series I noticed that in that way I wasn't taking into account the derivatives in terms of $x$.
Then I tried the following
$$x+\sqrt{x^2+1} = x + 1 + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}x^{2k} + o(x^{2n})$$
If I put
$$t' = x + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}x^{2k} + o(x^{2n})$$
the logarithm takes the form $\ln(1+t')$, but if I tried to expand this in Maclaurin Series I'll get a quite messy expression in powers of $t'$, which doesn't allow me to write the required expression in closed form (or maybe I don't have the knowledge to do that from it).
Hint
If we put $f(x)=\ln(x+\sqrt{1+x^2})$
then
$$f'(x)=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$$
$$=(1+x^2)^{-\frac 12}$$
$$=1-\frac 12 x^2+\frac 38 x^4-....$$
and integrate to get
$$f(x)=x-\frac 16 x^3+\frac{3}{40} x^5-...$$