Prove that $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0$ (without trigonometric substitution)

Notice that by the substitution $x = 2 + u$,

$$ I = \int_{-2}^{2} \frac{\log(2 + u)}{\sqrt{4 - u^2}} \, du = \int_{0}^{2} \frac{\log(4 - u^2)}{\sqrt{4 - u^2}} \, du. $$

On the other hand, by the substitution $x = 4 - v^2$ (or equivalently $v = \sqrt{4 - x}$), we have

$$ I = \int_{0}^{2} \frac{\log(4 - v^2)}{v \sqrt{4 - v^2}} \cdot 2v \, dv = 2 \int_{0}^{2} \frac{\log(4 - v^2)}{\sqrt{4 - v^2}} \, dv. $$

Comparing two formulas give $I = 2I$ and therefore $I = 0$.