A Treatise on The Integral Calculus

$I = \int_0^4 \frac{ln(x)}{\sqrt{4x-x^2}} dx = \int_0^4 \frac{ln(x)}{\sqrt{4 - (x - 2)^2}} dx$

Substitute $x - 2 = 2cos\theta$

$I = \int_0^{\pi} ln(2+2cos\theta) d\theta = \int_0^{\pi} ln(2+2cos(\pi-\theta)) d\theta = \int_0^{\pi} ln(2-2cos\theta) d\theta$

Adding first and last terms in the above equivalence,

$2I = \int_0^{\pi} ln(4-4cos^2\theta) d\theta = \int_0^{\pi} ln(4sin^2\theta) d\theta = 2\int_0^{\pi} ln(2sin\theta) d\theta$

$\Rightarrow I = \int_0^{\pi} ln(2sin\theta) d\theta = 2\int_0^{\pi/2} ln(2sin\theta) d\theta = \pi ln2 + 2\int_0^{\pi/2}ln(sin\theta) d\theta$ Call this equation $A$

Let $I_1 = \int_0^{\pi/2}ln(sin\theta) d\theta$

$I_1 = \int_0^{\pi/2}ln(cos\theta) d\theta$

Adding the above two equations,

$2I_1 = \int_0^{\pi/2}ln(sin\theta cos\theta) d\theta = \int_0^{\pi/2}ln(\frac{sin2\theta}{2}) d\theta$

$2I_1 = \int_0^{\pi/2}ln(sin2\theta) d\theta - \frac{\pi}{2}ln(2)$ Call this equation $B$

For the integral term on the right, substitute $2\theta = \alpha$

$\int_0^{\pi/2}ln(sin2\theta) d\theta = \frac{1}{2}\int_0^{\pi}ln(sin \alpha) d\alpha = \frac{2}{2}\int_0^{\pi/2}ln(sin\alpha)d\alpha$ which is same as $I_1$

Using the above result in equation $B$ , we get

$2I_1 = I_1 - \frac{\pi}{2}ln(2) \Rightarrow I_1 = -\frac{\pi}{2}ln(2)$

Using the above result in equation $A$ , we get

$I = \pi ln2 + 2\left( -\frac{\pi}{2}ln(2)\right) = 0$

Some identities used in the above solution:

$\int_0^{a} f(x)dx = \int_0^a f(a-x)dx$

$\int_0^{2a} f(x)dx = 2\int_0^a f(x)dx$ if $f(x) = f(2a - x)$

$S= \int_0^4 \dfrac{\ln(x)}{\sqrt{4x-x^2}}=\int_0^4 \dfrac{\ln(4-x)}{\sqrt{4x-x^2}} \to 2S=\int_0^4 \dfrac{\ln(4x-x^2)}{\sqrt{4x-x^2}}$ let $I(a)=\int_0^4 (4x-x^2)^a dx=4\int_0^1 16^a x^a(1-x)^a dx= 4*16^a \beta(a+1,a+1)$ then $I'(a)= 4\ln(16)16^a \beta(a+1,a+1)+8*16^a \beta(a+1,a+1)(\psi(a+1)-\psi(2a+2))$ putting a=-1/2 we have $2S=I'(-1/2)=\ln(16)\pi +2\pi(-\gamma-2\ln(2)+\gamma)=0\to S=\boxed{0}$