Finding the value of $\sum_{k=1}^{\infty} \frac{x^{4k-3}}{4k-3}$ 
This is my power series: $$\sum_{k=1}^{\infty} \frac{x^{4k-3}}{4k-3}$$

I need to find the sum of it. Unfortunately, I have kind of no idea how to do it. I think I need to substitute with something. One of the series that I do know of is $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ which looks similar, $(4k-3)$ instead of $k$ but I really don't know how to substitute them. 
Maybe I can do this, which will help, but I'm not sure?
$$\sum_{k=1}^{\infty} \frac{x^{4k-3}}{4k-3} = \sum_{k=1}^{\infty} \int_0^x \frac{x^{4k-4}}{4k-4} dx$$
Edit: I can see how I made a mistake with the integral.
At least with that I could have something like ${(x^4)}^{k-1}$? Any ideas?

$$ \frac{x^{4k-3}}{4k-3} = \int_{0}^{x} z^{4k-4}\,dz $$
hence:
$$ \sum_{k\geq 1}\frac{x^{4k-3}}{4k-3} = \int_{0}^{x}\sum_{k\geq 1} z^{4k-4}\,dz = \int_{0}^{x}\frac{dz}{1-z^4} $$
but:
$$ \frac{1}{1-z^4} = \frac{1}{2}\left(\frac{1}{1+z^2}+\frac{1}{1-z^2}\right) $$
so:

$$ \sum_{k\geq 1}\frac{x^{4k-3}}{4k-3} = \frac{1}{2}\,\arctan(x)+\frac{1}{4}\,\log\left(\frac{1+x}{1-x}\right)$$
  as soon as $|x|<1$.

