We have the following identity (see Bateman, H. (1953). Higher Transcendental Functions [Volumes I], p. 25.) $$(*)\quad \Gamma(s)\, \zeta(s,\nu) = \int_{0}^{1} x^{\nu-1} \,(1-x)^{-1} \Bigr(\log 1/x\Big)^{s-1} \, dx; \quad \Re e (s)>1,\Re e (\nu)>0,$$ where $\Gamma(s)$ is the Gamma function and $\zeta(s,\nu)$ is the generalized zeta function http://mathworld.wolfram.com/HurwitzZetaFunction.html
Now, I would like compute the following $$I_{\alpha,\beta} = \int_{0}^{1} x^{\alpha} \,(1-x)^{-2} \Bigr(\log 1/x\Big)^{\beta} \, dx; \quad \alpha>0,\, -1<\beta<0.$$ Thank you in advance
Notice: \begin{aligned} \frac{d}{dx} \left[x^a(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\right] &= ax^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta} \\ &\quad + x^a(1-x)^{-2}\left(\log \frac{1}{x} \right)^{\beta} \\ &\quad + \beta x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta-1} \end{aligned} and hence \begin{aligned} \int_0^1 x^a(1-x)^{-2}\left(\log \frac{1}{x} \right)^{\beta} \,dx &= \int_0^1 \frac{d}{dx} \left[x^a(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\right]\, dx \\ &\quad - a\int_0^1 x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta}\, dx \\ &\quad - \beta \int_0^1 x^{a-1}(1-x)^{-1}\left(\log \frac{1}{x} \right)^{\beta-1} \, dx \end{aligned} It seems to me that the right hand side of this last equality is solvable (either by the fundamental theorem of calc, or by your identity above).