Turning sum to integral representation I'd like to turn this sum: 
\begin{align}\sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)} \end{align}
into an integral $\displaystyle \int_{a}^{b} g(x) \space dx$.
There seems to be many methods to either change or approximate sums as integrals. So I've become confused which approach would work.
In Is it possible to write a sum as an integral to solve it? robjohn used $\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n$ which looks similar to a Laplace Transforms.
I can't see how he gets rid of the n's so I'm not able to apply it here otherwise it seems promising.  But looking elsewhere there are also approximations methods such as: Turning infinite sum into integral which even more obscure at least to me.
How do I convert this sum to an integral?

Recall that
$$\frac{x^{n+1}}{3^{n+1}(n+1)} = \int_0^{x/3} t^n \ \mathrm{d}t$$
Hence
$$\sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)} = \sum_{n=0}^{\infty} \int_0^{x/3} t^n \ \mathrm{d}t = \int_0^{x/3} \sum_{n=0}^{\infty} t^n \ \mathrm{d}t = \int_0^{x/3} \frac{1}{1-t} \mathrm{d}t = -\ln \left( \frac{3-x}{3} \right)$$
This holds for $|x|<3$.
