Evaluating the integral $\int_{-1}^1 \frac{\ln|z-x|}{\sqrt{1-x^2}}\mathrm dx$

Question

I don't know how to deal with this integral:

$$\int_{-1}^{1}{\ln\left(\,\left\vert\,z - x\,\right\vert\,\right)\over \,\sqrt{\vphantom{\large A}\,1 - x^{2}\,}\,}\,{\rm d}x\,,$$ where $z$ is a complex number.

Answer 1

Let $f(z)=\int_{-1}^1\dfrac{1}{\sqrt{1-x^2}}\ln|z-x|~dx$ ,

Then $\dfrac{df(z)}{dz}=\int_{-1}^1\dfrac{1}{(z-x)\sqrt{1-x^2}}dx$

According to http://www.wolframalpha.com/input/?i=int1%2F%28%28z-x%29%281-x%5E2%29%5E%281%2F2%29%29%2Cx%2C-1%2C1,

$\dfrac{df(z)}{dz}=\dfrac{\pi}{\sqrt{z^2-1}}$

$f(z)=\int\dfrac{\pi}{\sqrt{z^2-1}}dz=\ln(z+\sqrt{z^2-1})+C$

But I don't know how to find $C$.