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 $I = 0$.

$\displaystyle K=\int_0^4 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$

$\displaystyle K=\int_0^4 \dfrac{\ln (4-x)}{\sqrt{4x-x^2}}~dx+\int_0^4 \dfrac{\ln x}{\sqrt{4x-x^2}}~dx$

By the change of variable $y=4-x$, it's readily seen that the two preceding integrals are equal.

$\displaystyle K=\int_0^2 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx+\int_2^4 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$

Perform the change of variable $y=\sqrt{4x-x^2}$ in both preceding integrals,

$\displaystyle K=2\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx+2\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx=4\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx$

Thus,

$\displaystyle K=2\int_0^2 \dfrac{\ln x}{\sqrt{1-\left(\tfrac{x}{2}\right)^2}}dx$

Perform the change of variable $y=\dfrac{x}{2}$,

$\displaystyle K=4\int_0^1 \dfrac{\ln(2x)}{\sqrt{1-x^2}}dx$

$\displaystyle K=4\int_0^1 \dfrac{\ln 2}{\sqrt{1-x^2}}dx+4\int_0^1 \dfrac{\ln x}{\sqrt{1-x^2}}dx$

Perform the change of variable $x=\sin y$ in both integrals,

$\displaystyle K=4\ln(2) \int_0^{\tfrac{\pi}{2}} \dfrac{\cos y}{\sqrt{1-(\sin y)^2}}dy+4\int_0^{\tfrac{\pi}{2}} \dfrac{\ln (\sin y)\cos y}{\sqrt{1-(\sin y)^2}}dy=2\pi\ln 2+4\int_0^{\tfrac{\pi}{2}} \ln (\sin y)dy$

It is well known that $\displaystyle \int_0^{\tfrac{\pi}{2}} \ln (\sin y)dy=-\dfrac{\pi \log 2}{2}$

Thus, $K=0.$

Finally we get:

$\displaystyle \int_0^4 \dfrac{\ln (4-x)}{\sqrt{4x-x^2}}~dx=\int_0^4 \dfrac{\ln x}{\sqrt{4x-x^2}}~dx=0$

PS: To be compliant with the question.

$\displaystyle \int_0^1 \dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{1}{2}\int_0^1 x^{\tfrac{1}{2}-1}(1-x)^{\tfrac{1}{2}-1}dx=\dfrac{\Gamma\left(\tfrac{1}{2}\right)^2}{\Gamma(1)}=\dfrac{\pi}{2}$

$\displaystyle \int_0^1 \dfrac{\ln x}{\sqrt{1-x^2}}dx=\dfrac{1}{4}\dfrac{\partial}{\partial s}\left[\int_0^1 x^s(1-x)^{\tfrac{1}{2}-1}dx\right]_{s=-\tfrac{1}{2}}=\dfrac{1}{4}\dfrac{\partial}{\partial s}\left[\dfrac{\Gamma(s+1)\Gamma\left(\tfrac{1}{2}\right)}{\Gamma\left(s+1+\tfrac{1}{2}\right)}\right]_{s=-\tfrac{1}{2}}=-\dfrac{\pi\ln2}{2}$