Think of a beautiful substitution! Maybe

$I(a) = \displaystyle \int_{0}^{1} \dfrac{x^{a}-x^{2}}{\ln(x)} dx$
$\dfrac{dI}{da} =\displaystyle \int_{0}^{1} \dfrac{\partial}{\partial a} \dfrac{x^{a}-x^{2}}{\ln(x)} dx = \displaystyle \int_{0}^{1} x^{a}dx = \dfrac{1}{a+1}$
$\dfrac{dI}{da} = \dfrac{1}{a+1}$
$I(a) = \ln(a+1) + c$
$I(2) = 0 \to c = -\ln(3)$
$I(a) = \ln\left(\dfrac{a+1}{3}\right)$
$I(3) = \ln\left(\dfrac{4}{3}\right)$