Let us state our claim:
$$\int_{-1}^1 \frac{1}{a+bz} \frac{1}{\sqrt{1-z^2}} \; dz
= \frac{\pi}{\sqrt{a^2-b^2}}.$$
To start we introduce a dogbone contour as our contour with two
circles at either end parameterized by $-1 + \varepsilon
\exp(i\theta)$ where $\psi \le \theta \le 2\pi-\psi$ and $1 +
\varepsilon \exp(i\theta)$ where $-\pi+\psi\le \theta \le
\pi-\psi$ Here $\psi$ is negligeably small and the ends of the
arcs are not closed so as to leave a small opening. We connect
the two terminal vertices above the real axis by a line, the same
for the two vertices below. These will produce multiples of the
desired integral in the limit.
We will use
$$f(z) = \frac{1}{a+bz}
\exp(-1/2 \times\mathrm{LogA}(1+z))
\exp(-1/2 \times\mathrm{LogB}(1-z)).$$
Here $\mathrm{LogA}$ denotes the branch of the logarithm where $0
\le \arg \mathrm{LogA} \lt 2\pi$ and $\mathrm{LogB}$ where $-\pi
\le \arg \mathrm{LogB} \lt \pi.$ The branch cut from
$\mathrm{LogA}$ includes the interval $[-1,1]$ on the real axis
which is inside the dogbone contour while for $x\gt 1$ both
branch cuts apply. In fact in the latter case they cancel and we
have continuity across $(1,\infty)$ and may derive analyticity by
Morera's theorem as explained at this MSE
link. The
pole at $z = -a/b$ is not on either cut with $a\gt b \gt 0.$
For the continuity the rational factor is obviously the same
above and below the cut, while for the two logarithmic factors we
get for $x\gt 1$ and above the cut $\mathrm{LogA}(1+x) =
\log(x+1)$ and $\mathrm{LogB}(1-x) = \log(x-1)-\pi i$ (rotation)
and below the cut $\mathrm{LogA}(1+x) = \log(x+1) + 2\pi i$ and
$\mathrm{LogB}(1-x) = \log(x-1) + \pi i$ (rotation again). This
yields above the cut
$$\exp(-1/2\times \log(x+1))\exp(-1/2\times(\log(x-1)-\pi i))
\\ = \frac{1}{\sqrt{x^2-1}} \exp(1/2 \times \pi i) =
\frac{i}{\sqrt{x^2-1}}$$
and below the cut
$$\exp(-1/2\times(\log(x+1)+2\pi i))\exp(-1/2\times(\log(x-1)+\pi i))
\\ = \frac{1}{\sqrt{x^2-1}} \exp(-1/2 \times 3\pi i) =
\frac{i}{\sqrt{x^2-1}}$$
and we have continuity across the cut. (For the cut itself the
values are from the above-the-cut case.) By Morera we have
analyticity in $\mathbb{C}\backslash [-1, 1]$ for the square root
term. The value may be simplified to $1/\sqrt{1-x^2}$ if
desired. Next we need to verify that the two segments above and
below the single branch cut enclosed by the dogbone contour
(i.e., the line connecting $-1$ to $1$) are multiples of the
target integral. We get above the cut
$$\exp(-1/2\times(\log(x+1))\exp(-1/2\times(\log(1-x))
= \frac{1}{\sqrt{1-x^2}}$$
and below
$$\exp(-1/2\times(\log(x+1)+2\pi i))\exp(-1/2\times(\log(1-x)))
= -\frac{1}{\sqrt{1-x^2}}.$$
This means that with a counter-clockwise traversal of the contour
we pick up the target integral times a factor of $-2.$ We will
compute the integral using the poles outside of the dogbone
contour. Next to compute the residues we have that the residue at
infinity is zero by inspection. We get for the pole at $-a/b$
that the residue is
$$\;\underset{z=-a/b}{\mathrm{res}}\; f(z) =
\;\underset{z=-a/b}{\mathrm{res}}\;
\frac{1}{b} \frac{1}{z+a/b} \frac{1}{\sqrt{1-z^2}} \\
= \frac{1}{b} \frac{1}{\sqrt{1-a^2/b^2}}
= \frac{1}{\sqrt{b^2-a^2}}.$$
Now the contour produces twice the desired value as explained earlier
and hence it is given by (poles outside rather than inside contour)
$$- \frac{1}{2} \times - 2\pi i \times
\frac{1}{\sqrt{b^2-a^2}}
= \frac{\pi i}{\sqrt{b^2-a^2}}$$
which is
$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{\sqrt{a^2-b^2}}.}$$
To be rigorous we also need to show that the contribution from
the circular arcs vanishes in the limit. We get for the left arc
with $z = -1 + \varepsilon \exp(i\theta)$ where $0\le\theta\lt
2\pi$ (the arc does not cross any cuts of the two logarithms):
$$\frac{1}{a-b+b\varepsilon\exp(i\theta)}
\\ \times \exp(-1/2\times \mathrm{LogA}(\varepsilon \exp(i\theta)))
\exp(-1/2\times \mathrm{LogB}(2-\varepsilon \exp(i\theta)))
\\ = \frac{1}{a-b+b\varepsilon\exp(i\theta)}
\\ \times \exp(-1/2\times (\log\varepsilon + i\theta))
\exp(-1/2\times \mathrm{LogB}(2-\varepsilon \exp(i\theta))).$$
The middle term is $\exp(- i\theta/2)
\frac{1}{\sqrt{\varepsilon}}$ and the other two are constant in
the asymptotic as $\varepsilon\rightarrow 0$. Multiply by $2\pi
\varepsilon$ to have it vanish. We get for the right arc with $z
= 1 + \varepsilon \exp(i\theta)$ where $-\pi\le\theta\lt\pi$
(this arc crosses both cuts, so we have to be careful):
$$\frac{1}{a+b+b\varepsilon\exp(i\theta)}
\\ \times \exp(-1/2\times \mathrm{LogA}(2+\varepsilon \exp(i\theta)))
\exp(-1/2\times \mathrm{LogB}(-\varepsilon \exp(i\theta))).$$
We may switch this to
$$\frac{1}{a+b+b\varepsilon\exp(i\theta)}
\\ \times \exp(-1/2\times \Re \mathrm{LogA}(2+\varepsilon \exp(i\theta)))
\exp(-1/2\times \Re \mathrm{LogB}(-\varepsilon \exp(i\theta))).$$
This is because the two imaginary components from the two
logarithms produce modulus one, being situated on the unit
circle. We obtain $\frac{1}{\sqrt{\varepsilon}}$ from the third
term and the other two are once more constant in the
asymptotic. Hence the contribution from this arc will vanish as
well on multiplication by $2\pi\varepsilon.$ These two arcs were
in fact indented by $\psi$ which we have going to zero.
This concludes the argument.
