Evaluate $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$ 
Evaluate
  $$\displaystyle\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$$

How do I evaluate this integral?  I know that the result is $0$, but I don't know how to obtain this.  Wolfram|Alpha yields a non-elementary antiderivative for the indefinite integral, so I don't think I can directly integrate and then plug in the upper/lower limits.

First let $t = x-2$ this way $4x-x^2 = 4 - (x-2)^2 = 4-t^2$. Substitute, 
$$ \int_{-2}^2 \frac{\log(t+2)}{\sqrt{4-t^2}} ~ dt $$
Now let, $\theta = \sin^{-1}\tfrac{t}{2}$ so that $2\sin \theta = t$ and hence, after substitute, 
$$ \int_{-\pi/2}^{\pi/2} \frac{\log [2(1+\sin \theta)]}{2\cos \theta} 2\cos \theta ~ d\theta = \pi \log 2 + \int_{-\pi/2}^{\pi/2} \log(1+\sin \theta)~d\theta  $$
To solve this integral, replace $\theta$ by $-\theta$,
$$ I = \int_{-\pi/2}^{\pi/2} \log(1+\sin \theta) ~d\theta= \int_{-\pi/2}^{\pi/2} \log(1-\sin \theta)~d\theta$$
Now, 
$$ I + I = \int_{-\pi/2}^{\pi/2} \log(1-\sin^2 \theta) ~ d\theta = 4\int_{0}^{\pi/2} \log (\cos \theta) ~ d\theta$$
The last integral is a well-known integral that computes to $-\frac{\pi}{2}\log 2$. 
Your final answer is, $\pi \log 2 -\pi\log 2$.
