Identity about integral of logarithm

We don't need integration by parts, indeed by $x^2=t \implies 2xdx=dt$ we have

$$\int_{0}^{\alpha}\frac{x}{\log(x)}dx=\int_{0}^{\alpha^2}\frac12\frac{1}{\log(\sqrt t)}dt$$

then recall that $\log A^n=n\log A$.

For a slightly different approach, suppose we define $$f(t) = \int_0^t \frac{x}{\ln(x)} \; dx$$ and $$g(t) = \int_0^t \frac{1}{\ln(x)} \; dx$$ Then $$f'(t) = \frac{t}{\ln(t)}$$ and $$g'(t^2) = \frac{1}{\ln(t^2)} \cdot 2t = \frac{2t}{2 \ln(t)} = \frac{t}{\ln(t)}$$ so $f'(t) = g'(t^2)$, from which we conclude that $f(t)-g(t^2) = C$, a constant. Since $f(0) = g(0)$, we conclude $C= 0$, hence $f(t) = g(t^2)$.