Prove equality $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$ $\sqrt{1+x^2}\cdot\ln (x+\sqrt{1+x^2}) = x+\frac{x^3}{3} - \frac{2}{3}\frac{x^5}{5}+ \frac{2}{3}\frac{4}{5}\frac{x^7}{7}-...$
I stuck with performing right part. Where are these coefficients from?
 A: We are looking for the Taylor series of 
$$f(x)=\sqrt{1+x^2}\int\frac{dx}{\sqrt{1+x^2}}\tag{1}$$
that is clearly related with the Taylor series of
$$ g(x) = \sqrt{1-x^2}\int\frac{dx}{\sqrt{1-x^2}} = \sqrt{1-x^2}\arcsin(x).\tag{2}$$
We may now invoke some heavy artillery. For instance, the following identity:
$$ \arcsin^2(x) = \sum_{n\geq 1}\frac{(4x^2)^n}{2n^2\binom{2n}{n}}\tag{3}$$
leading to:
$$ \frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n\geq 1}\frac{4^n}{(2n)\binom{2n}{n}} x^{2n-1} \tag{4} $$
and to:
$$ x-\sqrt{1-x^2}\arcsin(x) = \int\frac{x\arcsin(x)}{\sqrt{1-x^2}}\,dx = \sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{5} $$
from which:
$$ g(x)=x-\sum_{n\geq 1}\frac{4^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}\tag{6}$$
and
$$\boxed{ f(x)=x-\sum_{n\geq 1}\frac{4^n(-1)^n}{(2n)(2n+1)\binom{2n}{n}}x^{2n+1}}\tag{7}$$
Two distinct prof of $(3)$ (one proof through Bonnet's recursion formula and Legendre polynomials, a second proof through the residue theorem) can be found in my course notes, at pages $18$ and $19$. A little warning: they are in Italian.
