I have been looking at the answers to the question - what is the value of $\int\_{0}^{1} \frac{dx}{\log x}(x^{\alpha} - x^{\beta})$. 
My question is, what has driven the questioners to seek the solution of the integral? I ask because I have been attempting some historical research on a paper of Euler - 'Analytical Speculations' where he uses this integral as a starting point to derive some other associated results. He produced a solution to this problem which is typically Eulerian.

First set $x^{\alpha} = e^{\alpha \log x}$. 
Use the series expansion for $e^{z}$ where $z = \alpha \log x $. Adopt a similar expansion for $x^{\beta}$.

As an aside, he develops the result that $ \int\_{0}^{1} (log x)^{n} dx = \pm n! $ positive when n is even, negative when odd.

Substituting the two series into the integral leads him to 

$\int\_{0}^{1} \frac{dx}{\log x}((x^{\alpha} - x^{\beta}) = \int\_{0}^{1} \frac{dx}{\log x}((1 + \alpha \log x + \frac{\alpha^{2} \log^{2} x}{1.2} + \frac{\alpha^{3} \log^{3} x}{1.2.3} + etc.) - (1 + \beta \log x + \frac{\beta^{2} \log^{2} x}{1.2} + \frac{\beta^{3} \log^{3} x}{1.2.3} + etc.)) $ 

which, integrating term by termsusing the above integral result, simplifies to

$(\alpha - \beta) - \frac{(\alpha^{2} - \beta^{2})}{2} + \frac{(\alpha^{3} - \beta^{3})}{3} - \frac{(\alpha^{4} - \beta^{4})}{4} +etc.$ which is the series for $\log(1 + \alpha) = \log(1 + \beta) = \log\frac{\alpha + 1}{\beta + 1}$.

As an additional 'extra', for little effort, he sets $\alpha = \alpha - 1$ and $\beta = \beta - 1$ in the original integral to derive
$\int\_{0}^{1} \frac{dx}{\log x}(x^{\alpha - 1} - x^{\beta - 1}) = \int\_{0}^{1} \frac{dx}{x\log x}(x^{\alpha} - x^{\beta}) = \log\frac{\alpha}{\beta}$.