This answer refers to the following
MSE post
and duplicate material has been omitted.
With these types of integrals usually what is being asked for is to
use two branches of the logarithm whose cuts cancel outside of the
integration interval.
Suppose we seek to compute
$$Q = \int_1^2 \frac{1}{x} \sqrt{(x-1)(2-x)} dx.$$
Re-write this as
$$\int_1^2 \frac{1}{z}
\exp(1/2\mathrm{LogA}(z-1))
\exp(1/2\mathrm{LogB}(2-z)) dz$$
and call the function $f(z).$
We must choose two branches of the logarithm $\mathrm{LogA}$ and
$\mathrm{LogB}$ so that the cut is on the real axis from $1$ to $2.$
This is accomplished when $\mathrm{LogA}$ has the cut on the negative
real axis and $\mathrm{LogB}$ on the positive real axis with the
arguments being $(-\pi,\pi)$ and $(0,2\pi).$
We use a dogbone contour which is traversed counterclockwise. Then
$\mathrm{LogA}$ gives the real value just above the cut but
$\mathrm{LogB}$ contributes a factor of $\exp(2\pi i \times 1/2).$
Below the cut $\mathrm{LogA}$ again produces the real value but so
does $\mathrm{LogB}.$ This implies that
$$Q (1 - \exp(2\pi i \times 1/2))
= - 2\pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$
or
$$Q = - \pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$
For the residue at zero we get
$$\exp(1/2\mathrm{LogA}(-1))
\exp(1/2\mathrm{LogB}(2))
\\ = \exp(1/2\times (-i\pi))\exp(1/2\log2)
= -i\sqrt{2}.$$
Now for the residue at infinity we use the formula
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right].$$
In the following we need to distinguish between the upper and the
lower half-plane. Assume $z=R e^{i\theta}$ with $0\le\theta\lt2\pi.$
Upper half-plane.
Here we have $$\mathrm{LogA}(1/z-1) =
\mathrm{LogA}(1-z) - \mathrm{LogA}(z)$$
and $$\mathrm{LogB}(2-1/z) =
\mathrm{LogB}(2z-1) - \mathrm{LogB}(z).$$
This gives for the function to evaluate the residue the term
$$- \frac{z}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(-1/2\mathrm{LogA}(z))\\ \times
\exp(1/2\mathrm{LogB}(2z-1))
\exp(-1/2\mathrm{LogB}(z)).$$
But we have in the upper half plane
$$\exp(-1/2\mathrm{LogA}(z))\exp(-1/2\mathrm{LogB}(z))
= \frac{1}{z},$$
so this becomes
$$- \frac{1}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(1/2\mathrm{LogB}(2z-1)).$$
Lower half-plane.
Here we have $$\mathrm{LogA}(1/z-1) =
\mathrm{LogA}(1-z) - \mathrm{LogA}(z)$$
and $$\mathrm{LogB}(2-1/z) = 2\pi i +
\mathrm{LogB}(2z-1) - \mathrm{LogB}(z).$$
This gives for the function to evaluate the residue the term
$$- \frac{z}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(-1/2\mathrm{LogA}(z))\\ \times
\exp(1/2\times 2\pi i)
\exp(1/2\mathrm{LogB}(2z-1))
\exp(-1/2\mathrm{LogB}(z)).$$
But we have in the lower half plane
$$\exp(-1/2\mathrm{LogA}(z))\exp(-1/2\mathrm{LogB}(z))
= -\frac{1}{z},$$
so this becomes
$$- \frac{1}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(1/2\mathrm{LogB}(2z-1)).$$
We have established matching terms for the upper and the lower
half plane.
The cut from the first term starts at one and extends to the right.
The cut from the second term starts at one half and also extends to
the right and we have cancelation of the overlapping segments for a
final cut being $[1/2, 1].$
Hence we certainly have analyticity of the exponential term in
a disk of radius one half round the origin.
We now evaluate the residue for this branch.
Here we have first that
$$\mathrm{LogA}(1-z) = - \mathrm{LogA}\frac{1}{1-z}$$
and second
$$\mathrm{LogB}(2z-1) = \pi i - \mathrm{LogB}\frac{1}{1-2z}.$$
where we choose $\pi i$ from the upper half plane to get analyticity
at the origin.
This finally yields
$$-\frac{1}{z^2} \exp(\pi i/2)
\exp\left(-\frac{1}{2}\mathrm{LogA}\frac{1}{1-z}\right)
\exp\left(-\frac{1}{2}\mathrm{LogB}\frac{1}{1-2z}\right).$$
We can extract coefficients from this either with the Newton binomial
or recognizing the mixed generating function of the unsigned Stirling
numbers of the first kind. Using the latter we find that
$$[z^n] \exp\left(u\log\frac{1}{1-z}\right)
= \frac{1}{n!} \times u(u+1)\cdots(u+n-1).$$
We need the first two terms of each. These are for the term
in $\mathrm{LogA}$
$$1 - \frac{1}{2} z +\cdots$$
and for the term in $\mathrm{LogB}$
$$1 - \frac{1}{2} 2z +\cdots$$
We have determined the residue at infinity of the original function to
be $$- \exp(\pi i/2) \times -\frac{3}{2} =
\frac{3}{2} i.$$
Recall that we had
$$Q = - \pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$
so substituting the computed values for the residues gives
$$-\pi i \left(-\sqrt{2}i + \frac{3}{2}i\right)
= \pi i \left(\sqrt{2}i - \frac{3}{2}i\right)
= \pi \left(-\sqrt{2} + \frac{3}{2}\right).$$
In order to be rigorous we also need to show continuity across the two
overlapping cuts on $(-\infty, 1)$ as shown in this MSE
link.
Remark. It really helps to think of the map from $z$ to $-z$ as a
$180$ degree rotation when one tries to visualize what is happening
here.
