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$