Solving definite integrals using complex integration Can anyone help with solving these two definite integrals using contour integration:
$$\int_{0}^{1}\frac{\sqrt{x(1-x)}}{(2x+1)^{2}(x+2)}dx$$
$$\int_{1}^{2}\frac{dx}{x\sqrt[7]{(x-1)^{3}(2-x)^{4}}}$$.
For the first one, I use a dog bone contour around $[0,1]$ on the real axis, and I have two poles at $-1/2$ and $-2$, the first one of the second order and the second is simple. At infinity the integral over the big circle is zero as R tends to infinity (Jordan's lemma). When I calculate the residues at the poles, I get $-\frac{2\sqrt{3}i}{3}$ for $-1/2$ and I believe I should get  $-\frac{\sqrt{3}i}{6}$, and for the second pole I get $\frac{\sqrt{6}i}{9}$, which I think is correct. Can anyone help me with finding the mistake in calculating the residue at the pole $-1/2$?
Similarly, for the second integral, I use a dog bone contour around $[1,2]$ on the real axis, and have only one pole at 0, so I believe that to get the solution I only have to calculate the residue at the pole 0, and I can not get the right solution. 
Can someone help me?

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
$2+\epsilon\exp(i\theta).$)
