Using the result,

For $I(\lambda ) = \int\limits_a^b {f(x,\lambda )dx} $, $\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_a^b {\left[ {\frac{\partial }{{\partial \lambda }}f(x,\lambda )} \right]dx} $

Let, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } – 1}}{{\ln x}}dx} $

$\frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\left[ {\frac{\partial }{{\partial \lambda }}\left( {\frac{{{x^\lambda } – 1}}{{\ln x}}} \right)} \right]dx} $

$ \Rightarrow \frac{{dI(\lambda )}}{{d\lambda }} = \int\limits_0^1 {\frac{{{x^\lambda }\ln x}}{{\ln x}}dx} $

$ = \left. {\frac{{{x^{\lambda + 1}}}}{{\lambda + 1}}} \right|_0^1 = \frac{1}{{\lambda + 1}}$

$I(\lambda ) = \int {\frac{1}{{\lambda + 1}}} d\lambda = \ln (\lambda + 1) + C$

$I(0) = C$

Also, $I(0) = \int\limits_0^1 {\frac{{{x^0} – 1}}{{\ln x}}dx} = 0$

So, $C = 0$

$\therefore I(\lambda ) = \ln (\lambda + 1)$

The given integral $ = I(e – 1) = \ln (e – 1 + 1) = 1$