<s>[INST] Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the next section. (Simmons' Calculus)
The hint given with the problem is to integrate another series. This series must be $$1-\frac{ 1}{2} x^2+\frac{ 1\cdot 3}{2\cdot 4} x^4-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^6+\cdots$$
But where does it come from? [How would I know to integrate that series, if it weren't for the hint?] 
The derivative of $\ln(x+\sqrt{1+x^2})$ is $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$, 
but I can't see how to get from $\displaystyle \frac{ 1}{\sqrt{1+x^2}}$ to that series.
The derivative can be written as $\displaystyle \frac{ 1}{x+\sqrt{1+x^2}}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$, but that looks even worse.
 [/INST] The first part of the derivate you calculated can be transformed to 
$$\frac{ 1}{x+\sqrt{1+x^2}} = 
\frac{ 1}{x+\sqrt{1+x^2}}  \frac{ x-\sqrt{1+x^2}}{x-\sqrt{1+x^2}} = 
\sqrt{1+x^2}-x$$ 
and therefore the derivate is 
$$\sqrt{x^2+1} - \frac{x^2}{\sqrt{x^2+1}}$$ 
The term $\frac{x}{\sqrt{x^2+1}}$ is the derivate of $\sqrt{x^2+1}$.
The expansion of 
$$\sqrt{1+x}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^n}$$
You can try to guess the coeffizients of the series:
Let $w(x)=\sqrt{x+1}$ Then 
$$(x+1)^{\frac{3}{2}}=w(x)(1+x)$$
Differentiating this equation gives
$$\frac{3}{2}w(x)=w(x)+(1+x)w'(x)$$
$$w(x)=2(x+1)w'(x)$$
If $p(x)=\sum_{n=0}^{\infty}a_n x^n$ then we get
$$a_n=2(na_n+(n+1)a_{n+1})$$
and
$$a_{n+1}=(-1)\frac{2n+1}{2(n+1)}a_n$$
if a powerseries $p(x)$ should fullfill this equation.
From this we get
$$a_n=\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}$$
by induction. So if w(x) there is a power series expansion of w(x) these are the coefficients. Both w(x) and the power series p(x) satisfy the equation 
$$\frac{f'(x)}{f(x)}=\frac{1}{x+1}$$
and therefore
$$(\log{(f(x)})'=\frac{1}{x+1}$$
Therefore 
$$(\log{(w(x)})'=(\log{(p(x)})'=\frac{1}{x+1}$$
$$\log{(w(x))}=\log{(p(x))}+C$$
$$w(x)=p(X)*D$$
where $D=e^C$. From $w(0)=p(0)=1$ we conclude that $D=1$ and $w(x)=p(x)$.
We have now proved that this is the power series expansion of $\sqrt{x+1}$.
Substitute $x$ by $x^2$ in $\sqrt{x+1}$ and you get $$\sqrt{1+x^2}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(1-2n)(n!)^2 4^n}x^{2n}}$$
differentiate the equation and get
$$\frac{x}{\sqrt{1+x^2}}=\sum_{n=1}^{\infty}{\frac{(-1)^n(2n)!(2n)}{(1-2n)(n!)^2 4^n}x^{2n-1}}$$
and therefore
$$\sqrt{x^2+1} - x*\frac{x}{\sqrt(1+x^2)}=\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n}x^{2n}}$$
This is the series of your hint.
Integration now gives
$$C+\sum_{n=0}^{\infty}{\frac{(-1)^n(2n)!}{(n!)^2 4^n (2n+1)}x^{2n+1}}$$
For x=0 the function is 0, therefore C=0
If you calculate $x$ from $y=\ln{(x+\sqrt{x^2+1})}$ you get $$x=\frac{(e^y)^2-1}{2e^y}=\frac{e^y-e^{-y}}{2}=\sinh(y)$$ Therefore your function is the inverse of the sinh, the arsinh.
You can chek your results at http://en.wikipedia.org/wiki/List_of_mathematical_series
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