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.

You can use the fact, as you write, that the derivative of such function is $$y' = \frac{1}{\sqrt{1+x^2}}$$
But from the binomial theorem you have that
$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {-\frac{1}{2} \choose k} x^{2k}$$
You can check by yourself that
$${-\frac{1}{2} \choose k} ={\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}}$$ 
using induction over the following expression
$$\frac{1}{{k!}}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)\left( { - \frac{5}{2}} \right) \cdots \left( {\frac{1}{2} - k} \right)$$
Where I simply replaced the value of  $n$ in $$\frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k + 1} \right)}}{{k!}}$$
As a consequence you get
$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} x^{2k}$$
Upon integration you have
$$ \log\left({x+\sqrt{1+x^2}}\right) = \sum_{k=0}^{\infty} {\left( { - 1} \right)^k}\frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} \frac{x^{2k+1}}{2k+1} = x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$
Remember that by definition $0!! = 1$ and $(-1)!! = 1$
