How to solve for a in the below equation:

I have the following equation

$$\int_0^1 \frac{a^x -1}{a-1} dx = r,$$

where the integral evaluates as:

$$\frac{1}{1-a} + \frac{1}{\log(a)} = r.$$

I would like to solve for a but this is proving to be quite complex. Any Suggestions?

There is no closed solution for $$a$$ in terms of "elementary" functions, but there is a solution with (generalized) Lambert W functions as follows:
Let $$a = e^{v}$$ then $$\frac{1}{1-e^{v}} + \frac{1}{v} = r$$ or $$e^{-v} = \frac{r}{r-1}\frac{v- 1/r}{ v- 1/(r-1)}$$ where this is of the general form
$$e^{-c v} = a_0 \frac{v- q}{ v- s}$$ with real constants $$c, a_0, q,s$$. This is a generalized form of the standard Lambert W function, a reference in wikipedia is here. So you can take it from there to assess solutions of this special form. The article "Maignan, Aude; Scott, T. C. (2016). "Fleshing out the Generalized Lambert W Function". SIGSAM. 50 (2): 45–60" cited in Wikipedia can be found here.