Integration of $\int_{-1}^{1}\frac{dx}{(x-a)\sqrt{1-x^2}}$ using Residue Calculus

Consider a contour that looks like this:

in the limit as $R$, the radius of $C_R$, goes to $\infty$. The contour should be constructed so that the point $z=a$ is inside the contour. The residue theorem then tells you $$\oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}=-\frac{2\pi i}{\sqrt{1-a^2}}=\frac{2\pi}{\sqrt{a^2-1}}$$ where $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a-1),\arg(a+1)\in [0,2\pi)$. You can show that the integral over $C_R$ goes to zero as $R\rightarrow \infty$. The integral over the segments of the contour parallel to the imaginary axis obviously cancel each other. And you stated that you have already shown that the integral over $C_\rho$ goes to zero as $\rho\rightarrow 0$.

So now we must deal with the straight segments on opposite sides of the branch cut. We selected the branch with $\arg(1+z)\in [0,2\pi)$, $\arg(1-z)\in [-\pi,\pi)$. The discontinuity over the branch cut is captured by $\arg(1+z)$. So parametrize the upper segment using $z = \xi - 1$ and the lower segment by $z = \xi e^{2\pi i}-1$. Then we have:

\begin{align} \oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}&=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}+\int_0^2\frac{e^{2\pi i}d\xi}{(\xi e^{2\pi i} -a-1)\sqrt{2-\xi e^{2\pi i}}\sqrt{\xi e^{2\pi i}}}\\ &=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}-\int_0^2\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}} \\ &=\int_1^{-1}\frac{dx}{(x-a)\sqrt{1-x^2}}-\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \qquad \text{(subst. }x=\xi-1\text{)} \\ &= -2\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \\ &= \frac{2\pi}{\sqrt{a^2-1}} \end{align} From this you can conclude that $$\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} = -\frac{\pi}{\sqrt{a^2-1}}$$ remembering that $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a+1),\arg(a-1) \in [0,2\pi)$.