I am trying to solve $$\int_{-1}^1\frac{\log(z-x)}{\sqrt{1-x^2}}dx$$ for z real. I would like a hint as to how to approach the integral (I took $\log(z)$ out and tried a trigonometric/hyperbolic substitution, but things don't seem to get nicer).
Update: From Gradshteyn one can find what the answer looks like for $0\leq 1\leq|z|$, namely $$\int_{-1}^1\frac{\log(z-x)}{\sqrt{1-x^2}}dx=\pi\log\frac{z+\sqrt{z^2-1}}{2}$$, but the references therein don't contain information on how to evaluate it.
For z complex (which I want to compute in the end) the answer should be
$$\int_{-1}^1\frac{\log|z-x|}{\sqrt{1-x^2}}dx=\pi\log\frac{|z+\sqrt{z^2-1|}}{2}$$
which I have no idea how to attack, hence I thought of dealing with the real case first