A few days ago I stumbled upon this integral $$ f(y,c) :=\int_{-c-\sqrt{2}}^{-c+\sqrt{2}}\sqrt{2-(x+c)^2}\; \ln\left|\frac{x+y}{x-y}\right| \;\mathrm{d}x $$ where $c>0$ and $y\in(-c-\sqrt{2},-c+\sqrt{2})$. After thinking for a while, it doesn't seem to be easy to solve (at least via standard tricks).
Thus, I wonder whether $f(y,c)$ admits a closed-form expression. Even an expression computed via symbolic math softwares would be fine (I don't currently have any of these softwares installed in my laptop...) Thanks for your help.
First generalise your integral to depend on $a$, and change variables so that $x \to a(x - c)$. Then $$f(y,c,a) := a^2\int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|\frac{x-\frac{c-y}{a}}{x-\frac{c+y}{a}}\right| \;\mathrm{d}x,$$ and choosing $a=\sqrt{2}$ reduces to your integral.
Now define $$g(A):= \int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|x-A\right| \;\mathrm{d}x.$$ Your original $f$ can then be expressed as $$f(y,c,a) = a^2\left(g\left(\frac{c-y}{a}\right) - g\left(\frac{c+y}{a}\right)\right),$$ so we've reduced to a one-dimensional problem. Given the absolute value in the log, there are different cases depending on whether $|A|>1$ or not.
When $|A|\ge1$ the $\ln(0)$ singularity at $x=A$ lies outside of the range of integration, and we have that $$g(A)= \int_{-1}^{1}\sqrt{1-x^2}\; \ln\left(x+|A|\right) \;\mathrm{d}x.$$ In this region Mathematica gives a solution in terms of a hypergeometric function, $$g(A) = -\frac{\pi}{16}\left(\frac{1}{A^2}\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};\frac{1}{A^2}\right) + 4\ln\left(\frac{1}{A^2}\right)\right).$$
We can write this expression in terms of elementary functions using the following identity for $\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};z\right)$, that $$z\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};z\right) = -8\left(\frac{1-\sqrt{1-z}}{z} + \mathrm{arctanh}\left(\sqrt{1-z}\right) -\ln(2\sqrt{e})\right) -4\ln z.$$Inserting this into the expression for $g(A)$ we find that$$g(A) = \frac{\pi}{2}\left(A^2 - \sqrt{A^2(A^2-1)} + \mathrm{arctanh}\left(\sqrt{1-\frac{1}{A^2}}\right) - \ln(2\sqrt{e})\right),$$ for $|A|\ge1$.
For $|A|<1$ there is a singularity in the middle of the range of integration. To treat this carefully I believe you'd have to write$$g(A)= \int_{-1}^{A}\sqrt{1-x^2}\; \ln\left(A-x\right) \;\mathrm{d}x + \int_{A}^{1}\sqrt{1-x^2}\; \ln\left(x-A \right) \;\mathrm{d}x$$ and take limits up to $A$ from the left in the first integral and from the right in the second. I don't know much about this, but an obvious question to ask as this point is; can we extend the expression for $g(A)$ when $|A|\ge1$ to $|A|<1$?
It's clear that this is not going to be true in the most naive way because $g(A)$ is real for $|A|<1$, but the expression for $|A|\ge1$ is complex when $|A|<1$, eg. from the $\sqrt{A^2-1}$. However, all is not lost; numerically in Mathematica I find that the integral matches up with the real part of the functional form I gave. So I have compelling numerical evidence that $$\int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|x-A\right| \;\mathrm{d}x = \frac{\pi}{2}\Re\left(A^2 - \sqrt{A^2(A^2-1)} + \mathrm{arctanh}\left(\sqrt{1-\frac{1}{A^2}}\right) - \ln(2\sqrt{e})\right)$$ for all $A\in\mathbb{R},$ solving your problem in terms of elementary functions for all regions of parameter space.