Reverse engineering a Taylor expansion 2 So there is the sum:
$$S(x) = \frac{x^3}{3(1!)} + \frac{x^6}{6(2!)} + \frac{x^9}{9(3!)} \text{ }...$$
and we are instructed to find the sum of the series in a small expression. 
I took the derivative to cancel out the increasing terms on the bottom:
$$S'(x) = \frac{x^2}{(1!)} + \frac{x^5}{(2!)} + \frac{x^8}{(3!)} \text{ }...$$
And then I multiplied by X on both sides, and replaced X with X^(1/3) and added 1 to both sides:
$$xS'(x^{1/3})+1 = 1+ \frac{x}{(1!)} + \frac{x^2}{(2!)} + \frac{x^3}{(3!)} \text{ }...$$
which is
$$xS'(x^{1/3})+1 = e^x $$
But when working my way back, I get
$$\int{S'(x^{1/3})} = \int{\frac{e^x-1}{x}} $$
which is impossible to integrate. Is there any way to work around this?

You properly identified that $$S'(x) = \frac{x^2}{1!} + \frac{x^5}{2!} + \frac{x^8}{3!}+ \cdots=\frac 1x\Big(\frac{x^3}{1!} + \frac{x^6}{2!} + \frac{x^9}{3!}+ \cdots\Big)=\frac{e^{x^3}-1}{x}$$ So, this write $$\frac{dS}{dx}=\frac{e^{x^3}-1}{x}$$ Now, changing variable $x=\sqrt[3]y$, this leads to $$\frac{dS}{dy}=\frac{e^y-1}{3 y}$$ and integration leads to $$S=\frac{1}{3} (\text{Ei}(y)-\log (y))+C$$ where appears the exponential integral $\text{Ei}$ which cannot be expressed in terms of any elementary functions and I suppose that this is the end of the story and $$S=\frac{1}{3}\text{Ei}\left(x^3\right)-\log (x)+C$$ The exponential integral can be expressed as incomplete gamma function but I do not think that this could help you in any manner.
