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.

$\int_{0}^{4}\frac{\ln{x}}{\sqrt{4x-x^2}}dx$
$=\int_{-2}^{2}\frac{\ln{(t+2)}}{\sqrt{4-t^2}}dt$
$=\frac12\int_{-2}^{2}\frac{\ln{(t+2)}}{\sqrt{4-t^2}}dt
+\frac12\int_{-2}^{2}\frac{\ln{(-t+2)}}{\sqrt{4-t^2}}dt$
$=\frac12\int_{-2}^{2}\frac{\ln{(4-t^2)}}{\sqrt{4-t^2}}dt$
$=0$
