$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x}\right)\mathrm dx=\gamma\tag1$$
Messing around with $(1)$ using wolfram integrator, we have
$$\int_{0}^{1}\left({x^u\over \ln(x)}+{x^v\over 1-x}\right)\mathrm dx=F(u,v)\tag2$$
$H_0=0.$
How do we show that $F(u,v)=\gamma-H_v+\ln(u+1)?$