Show that $\int_0^\infty \frac{\log x}{\prod_{u=0}^n (1+\alpha_u\,x)}\,dx = ...$

Question

Prove that, by complex analysis, $$\int_0^\infty \frac{\log x}{\prod_{u=0}^n (1+\alpha_u\,x)}\,dx = \frac{1}{n! \,2^{n+1}}\sum_{m=0}^n (-1)^{m+1}\binom{n}{m}\alpha_m^{n-1}\log^2(\alpha_m).$$ The real analysis approach is fairly straightforward, but rather tedious. As for complex analysis, I am unsure. My approach obviously fails: Begin by letting $$\oint_C f(z)\,dz$$ be a contour integral over a semi-circular arc where the branch along the negative real axis of the logarithm is chosen (perhaps this is the issue) and the contour is deformed into a keyhole contour with the orgin cut out. Then, by Cauchy's theorem, one can show a similar relation with the result. The issue lies when we integrate on the upper and lower branches, we get a $-x$ term on the denominator. How can I traverse this issue? At first I thought it is obvious to instead choose a different branch of the logarithm but I do not know how this would work. A hint is very much appreciated. Thank you!