Branch cuts for the contour integral $\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx$ How can the branch cut be handled in the contour integral, for $|b| \leq 1, \, a > 1$,
$$\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx \quad ?$$
If $a=1$ can the value of the integral be shown to be zero?

Edit: The contour involves a branch cut from $-a + \sqrt{a^{2}-1}$ to $\infty$ along the $x$-axis, a branch cut from $- a - \sqrt{a^{2}-1}$ to $- \infty$ along the $x$-axis. There are two poles $- z_{b} = e^{i(\theta_{b} - \pi)}$ and $- \frac{1}{z_{b}} = e^{-i(\theta_{b}+\pi)}$ where $\theta_{b} = \cos^{-1}(b)$. The resulting proposed value is:
\begin{align}
\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx = \frac{2 \pi}{\sqrt{1-b^{2}}} \, \tan^{-1}\left[ \frac{\sqrt{1-b^{2}} \, (a - \sqrt{a^{2}-1})}{1 - b \, (a - \sqrt{a^{2}-1})} \right].
\end{align}

The problem of the integrand is that, if we use the principal branches then the singular points somehow overlap, which makes it hard to deal with directly. So if you want to apply complex analysis technique, we need some modification.
Notice that the integral
$$ I = \mathrm{PV}\!\int_{-1}^{1} \frac{\log(x+a)}{(x+b)\sqrt{1-x^2}} \, dx $$
exists in principal value sense, due to the presence of the pole at $x = -b$. One way to circumvent this problem is to move the pole slightly away from the line of integration. So we introduce the following function:
$$ I(w) = \int_{-1}^{1} \frac{\log(x+a)}{(x+w)\sqrt{1-x^2}} \, dx. $$
Then we can express $I$ as
$$ I = \lim_{s \downarrow 0} \frac{I(b+is) + I(b-is)}{2}. \tag{1} $$
So it suffices to deal with $I(w)$ when $w$ lies outside the line $[-1, 1]$. Now choose a contour $C$ that winds $[-1, 1]$ counter-clockwise:

Using the principal square root, introduce the function $f(z) = i\sqrt{z-1}\sqrt{z+1}$. It is easy to see that the branch cut of $f$ is $[-1, 1]$ and that for $-1 < r < 1$,
$$ f(r + 0^+i) = -\sqrt{1-r^2}, \qquad f(r - 0^+i) = \sqrt{1-r^2}. $$
Then the integral is written as
$$ I(w) = \frac{1}{2} \oint_C \frac{\log(z+a)}{(z+w)f(z)} \, dz. $$
Now deform this contour as follows:

This can be done by enlarging the contour $C$ while wrapping around the singular points. (And it is possible since the integrand grows like $\mathcal{O}(\log R/ R^2)$ as the radius $R \to \infty$.) Then it follows that
\begin{align*}
I(w)
&= \pi i \int_{-\infty}^{-a} \frac{dx}{(x+w)f(x)} - \pi i \underset{z=-w}{\operatorname{Res}} \frac{\log(z+a)}{(z+w)f(z)} \\
&= \pi \int_{a}^{\infty} \frac{dx}{(x-w)\sqrt{x^2 - 1}} - \frac{\pi \log(a-w)}{\sqrt{-w-1}\sqrt{-w+1}}
\end{align*}
Plugging this to (1), we have
\begin{align*}
I &= \pi \int_{a}^{\infty} \frac{dx}{(x-b)\sqrt{x^2 - 1}} \\
&= \frac{2\pi}{\sqrt{1-b^2}} \left( \arctan\sqrt{\frac{1+b}{1-b}} - \arctan\sqrt{\frac{1+b}{1-b}\cdot\frac{a-1}{a+1}} \right).
\end{align*}
