Consider the integral I(α)=∫01xα−1lnxdxI(\alpha) = \int_0^1 \frac{x^\alpha - 1}{\ln x} dxI(α)=∫01lnxxα−1dx for α>−1\alpha > -1α>−1. Applying Leibniz's Rule, what is the value of I(α)I(\alpha)I(α)?
α\alphaα
ln(α+1)\ln(\alpha + 1)ln(α+1)
ln(α)\ln(\alpha)ln(α)
eαe^\alphaeα