Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $ 
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$

I was able to notice that:  $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx  $$
Where should I take it from here? (assuming I'm not the right way)
EDIT
Following the anwer:
$$\int \sum_{n=0}^\infty x^{2n-1} dx = \int \frac{\sum_{n=0}^\infty (x^2)^n}{x} dx = \int \frac{\frac{1}{1-x^2}}{x} dx= \int \frac{1}{x(1-x^2)}dx = \int \frac{1-x^2+x^2}{x(1-x^2)} = \int \frac{1}{x} dx + \int \frac{x}{1-x^2}dx = \ln(x) + \int \frac{1}{2}\frac{1}{1-t}dt + C = \ln(x) - \frac{\ln(x^2)}{2} + C$$  


*

*Is that right?

*How to evaluate $C$?

 A: Consider,
$$\sum_{n\ge 1} x^{n-1} = \frac{1}{1-x}$$
Let $x \rightarrow x^2$ and then integrate. 
A: $$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \\
\log \frac{1}{1-x} = x + \frac{x^2}{2} + \frac{x^3}{3} +\frac{x^4}{4} + \cdots $$
and so 
$$\frac{1}{2}(\, \log(1+x) + \log \frac{1}{1-x}\,) = x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots$$
or
$$ x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots= \log \sqrt{ \frac{1+x}{1-x}}$$
for $|x| \le 1$, $x \ne \pm 1$.
$\tiny{\log \sqrt{ \frac{1+x}{1-x}}= \text{arctanh} x= \frac{\arctan(ix)}{i}}$
A: You got the index wrong. The actual sum is 
$$ \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int \sum_{n = 0}^{\infty} x^{2n} \,dx = \int \frac{1}{1-x^2}\,dx $$
This sum converges for $|x| < 1$ and the integral evaluates to
$$ f(x) = \frac{1}{2}\, \ln \left| \frac{1+x}{1-x} \right| + C $$
The constant $C$ can be found by looking at the series expansion (what's $f(0)$?)
A: Use only your first identity. You should check where the series is absolutelly convergent. In that interval you can change the integral with the sum. That will lead to a known series you shall be able to evaluate. (If you can't solve with the hint I edit this answer later).
