Using Leibnitz Theorem,
$\frac{\mathrm{d}I }{\mathrm{d} a}$= $\frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1} \frac{x^{\alpha }-1}{log x}dx$
= $\int_{0}^{1}\frac{\partial }{\partial a} \frac{x^{\alpha }-1}{log x}dx$ (differentiation of Integral function is equal partial differentiation of that function)
= $\int_{0}^{1} \frac{x^{a} log x}{log x}$
= $\int_{0}^{1} x^{a} dx$
= $\frac{x^{a+1}}{a+1}$
= $\frac{{1}}{a+1}$
$\int dI$ =$\int \frac{1}{a+1} da$
= ln(a+1) + c