Solution:

$\begin{array}{ll}{\text { let }} {I=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x} \\ {\text { Taking 'a' as parameter, }}\end{array}$

$I(a)=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x---------(1)$

differentiate w.r.t a

$\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{1} \frac{x^{a}-1}{\log x} d x$

$\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \frac{\partial}{\partial a} \frac{x^{a}-1}{\log x} d x$

$\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \frac{x^{a} \cdot \log x}{\log x} d x$

$\therefore \frac{d I(a)}{d a}=\int_{0}^{1} x^{a} d x$

$\therefore \frac{d I(a)}{d a}=\left[\frac{x^{a+1}}{a+1}\right]_{0}^{1}$

$\therefore \frac{d I(a)}{d a}=\frac{1}{a+1}-0$

$\therefore \frac{d I(a)}{d a}=\frac{1}{a+1}$

$Now,$ integrate w.r.t a,

$\begin{aligned} \mathrm{I}(\mathrm{a}) =\int \frac{1}{a+1} d a \\ \mathrm{I}(\mathrm{a})=\log (\mathrm{a}+1)+\mathrm{c} --------(2)\\ \text { where } \mathrm{c} \text { is constant of integration } \end{aligned}$

put $a=0$ in eqn $(1)$

$\quad I(0)=\int_{0}^{1} 0 d x=0$

And

From eqn $(2), \quad I(0)=c$

$\therefore c=0$

$\therefore I=\log (a+1)$

Hence Proved