This question already has an answer here:
Here’s a method using the Feynman integration trick:
Consider
$$I(a)=\int\limits_0^1 \frac{x^a-1}{\log x}\,dx$$
Then we have
$$I'(a)=\frac{d}{da}\int\limits_0^1 \frac{x^a-1}{\log x}\,dx=\int\limits_0^1 \frac{d}{da}\left[\frac{x^a-1}{\log x}\right]\,dx=\int\limits_0^1 x^a\,dx=\frac{1}{a+1}$$
Now integrating with respect to $a$ gives
$$I(a)=\log(a+1)+C$$
Now, to find $C$, we can note that the $I(0)=0$ (because when $a=0$ the numerator turns into $1-1=0$). Hence $0=\log(0+1)+C$, so $C=0$.
Hence we have $I(a)=\log(a+1)$, and in particular we care about the $a=1$ case in which the solution is just $\log(2)$.