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?

Outline: Let 
$$f(t)=\sum_1^\infty t^{4k-4}.$$  The  power series we were given can be obtained by integrating the above series term by term from $0$ to $x$.
But $f(t)$ is a geometric series with common ratio $t^4$, and therefore has sum $\frac{1}{1-t^4}$. 
Now carry out the integration (partial fractions).
