For the second one, use a dogbone contour traversed counterclockwise
around $z=1$ and $z=2$ with the function
$$f(z) = \frac{1}{z}
\exp((-3/7)\mathrm{LogA}(z-1))
\exp((-4/7)\mathrm{LogB}(2-z))$$
with $\mathrm{LogA}$ having argument between $[0,2\pi)$ and branch cut
on the positive real axis and $\mathrm{LogB}$ having argument between
$[-\pi, \pi)$ and branch cut on the negative real axis. We claim the
branch cut of $f(z)$ is the interval $[1,2]$ with continuity across
the cut on $(2,\infty)$ (and hence analyticity by Morera's theorem).
For the first interval put $x=1+t+i\epsilon$ with $\epsilon\ge 0$ and
$t\in(0,1)$ so we are above the cut from the first logarithm. We are
in the half plane not containing the cut of the second logarithm. We
obtain
$$\frac{1}{x}
\exp((-3/7)\log(x-1))
\exp((-4/7)\log(2-x))
\\ = \frac{1}{x\sqrt[7]{(x-1)^3(2-x)^4}}.$$
Next put $x=1+t-i\epsilon.$ We are now below the cut from the first
logarithm. The cut of the second logarithm does not participate. We
get
$$\frac{1}{x}
\exp((-3/7)\log(x-1)+(-3/7)\times 2\pi i)
\exp((-4/7)\log(2-x))
\\ = \frac{\exp((-6/7)\pi i)}{x\sqrt[7]{(x-1)^3(2-x)^4}}.$$
The conclusion is that with
$$J = \int_1^2 \frac{1}{x\sqrt[7]{(x-1)^3(2-x)^4}} \; dx$$
the two straight segments from the dogbone contribute
$$(-1 + \exp((-6/7)\pi i)) J.$$
To see that we have continutity across $(2,\infty)$ we first put $x =
2 + t + i\epsilon$ so we are above the cut of $\mathrm{LogA}.$ We are
now below the cut of $\mathrm{LogB}.$ Combine to get
$$\frac{1}{x}
\exp((-3/7)\log(x-1))
\exp((-4/7)\log(x-2)+(-4/7)\times -\pi i)
\\ = \frac{\exp((4/7)\pi i)}{x\sqrt[7]{(x-1)^3(x-2)^4}}.$$
Next put $x = 2 + t -i\epsilon$ so we are below the cut of
$\mathrm{LogA}.$ We are now above the cut of $\mathrm{LogB}.$ Combine
to get
$$\frac{1}{x}
\exp((-3/7)\log(x-1)+(-3/7)\times 2\pi i)
\exp((-4/7)\log(x-2)+(-4/7)\times \pi i)
\\ = \frac{\exp((-10/7)\pi i)}{x\sqrt[7]{(x-1)^3(x-2)^4}}.$$
Seeing that $\exp((-10/7)\pi i) = \exp((4/7)\pi i)$ we indeed have
continuity across the cut as claimed. Observe also that
$\lim_{R\to\infty} 2\pi R \times 1/R/\sqrt[7]{R^3\times R^4} = 0$ so
the residue at infinity is zero. This leaves the residue at zero which
is particularly simple, we have
$$\left.\exp((-3/7)\mathrm{LogA}(z-1))
\exp((-4/7)\mathrm{LogB}(2-z))\right|_{z=0}
\\ = \exp((-3/7)\mathrm{LogA}(-1))
\exp((-4/7)\mathrm{LogB}(2))
= \exp((-3/7)\pi i) 2^{-4/7}.$$
We then have for our integral that
$$(-1 + \exp((-6/7)\pi i)) J = - 2\pi i
\exp((-3/7)\pi i) 2^{-4/7}$$
so that
$$J = - 2\pi i
\frac{\exp((-3/7)\pi i)}{-1+\exp((-6/7)\pi i)} 2^{-4/7}
\\ = - 2\pi i
\frac{1}{-\exp((3/7)\pi i)+\exp((-3/7)\pi i)} 2^{-4/7}.$$
Hence the desired answer is
$$\bbox[5px,border:2px solid #00A000]{
J = \frac{\pi}{\sin(3\pi/7)} 2^{-4/7}.}$$
For the two circular components of the dogbone we note that
$\lim_{\epsilon\to 0} 2\pi \epsilon \times 1/1/\epsilon^{3/7}/1^{4/7}
= 0$ and $\lim_{\epsilon\to 0} 2\pi \epsilon \times
1/2/1^{3/7}/\epsilon^{4/7} = 0$ so these two really do vanish.
(Parameterize by $1+\epsilon \exp(i\theta)$ and
$2+\epsilon\exp(i\theta).$)
