Femibern

$\large \displaystyle I(a) = \int_0^1 \frac{x^a -1}{\ln(x)} dx$

$\large \displaystyle \frac{dI(a)}{da} = \int_0^1 \frac{x^a \ln(x)}{\ln(x)} dx$

$\large \displaystyle \frac{dI(a)}{da} = \int_0^1 x^a dx$

$\large \displaystyle \frac{dI(a)}{da} = \frac{1}{a+1}$

$\color{#20A900} \boxed{ \large \displaystyle I(a) = \ln(a+1) + C}$

When $a = 0$ :

$\large \displaystyle I(0) = \ln(1) + C = \int_0^1 \frac{x^0 -1}{\ln(x)} dx = 0$

$\color{#20A900} \boxed{ \large \displaystyle C = 0}$

Thus:

$\color{#20A900} \boxed{ \large \displaystyle I(a) = \ln(a+1)}$

$\color{#20A900} \boxed{ \large \displaystyle s = I(2) = \ln(3)}$

So:

$\color{#3D99F6} \boxed{\large \displaystyle e^s = 3}$

It should be clear that $\int_0^1 \frac{x^b-1}{\ln (x)} dx = \ln (b+1)$ converges for $b>-1$ . In particular, $\int_0^1 \frac{x^2-1}{\ln (x)} dx = \ln (3)$ which leads to the result.