Calculate the sum $\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1}$ I want to calculate the sum : $$\sum_{n=0}^{\infty}\frac{x^{3n+1}}{3n+1} $$
I differentiated it, so $$\sum_{n=1}^{\infty}x^{3n}$$
by putting $ x = \sqrt[3]{t}$ it becomes :
$$\sum_{n=0}^{\infty}t^n$$
so $$f'(x) = \frac{1}{1-x^3}$$
by integrating it I should get the sum right ?
Thanks in advance! 

\begin{eqnarray*}
S=\sum_{n=0}^{\infty} \frac{x^{3n+1}}{3n+1} = \sum_{n=0}^{\infty} \int_{0}^x x^{3n} dx = \int_{0}^x \frac{dx}{1-x^3}
\end{eqnarray*}
Now do partial fractions
\begin{eqnarray*}
\frac{1}{1-x^3}= \frac{x+2}{3(1+x+x^2)}-\frac{1}{3(x-1)}
\end{eqnarray*}
Integrating this gives
\begin{eqnarray*}
S=\frac{1}{6} \ln(1+x+x^2)+\frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2x+1}{\sqrt{3}} \right)-\frac{\pi}{6 \sqrt{3}} +\frac{1}{3} \ln(1-x)
\end{eqnarray*}
