How to integrate $ \int_\gamma \left(\sum_{n=0}^{10} \frac{-ydx + (x-2n-1)dy}{(x-2n-1)^2+y^2} \right) $ I want to solve the following problem 

Let $\gamma$ be the polar curve $r=2+4\cos(\theta)$, $0\leq \theta \leq 2\pi$. Compute the integral
  \begin{align}
  \int_\gamma \left(\sum_{n=0}^{10} \frac{-ydx + (x-2n-1)dy}{(x-2n-1)^2+y^2} \right)  
\end{align}

For circle $r=1$ with $0\leq \theta \leq 2\pi$, with given $\int_\gamma \frac{-ydx + xdy}{x^2+y^2}$ I know how to do this integral. 
i.e., by putting $x=r\cos(\theta), y=r\sin(\theta)$ then 
\begin{align}
\int_0^{2\pi} \left( \sin(\theta)^2 + \cos(\theta)^2 \right) d\theta = 2 \pi 
\end{align}
How about that particular problem?   It seems I have difficulty for parameterizing $r=2+4\cos(\theta)$ in cartesian coordinates, and even so I don't understand the purpose of $(x-2n-1)^2$.  
 A: This is a test of how well you understand the concepts in vector calculus, not the mechanics. This particular vector field
$$\frac{\langle -y,x \rangle}{x^2+y^2}$$
is the most famous example of a curl-free but not conservative vector field. The vector field is not conservative on all of $\mathbb{R}^2$ because as you correctly noted it is not path independent. But on any simply connected subset of $\mathbb{R}^{2}$ not containing the origin, this vector field is conservative because as we can note
$$\frac{\langle -y,x \rangle}{x^2+y^2} = \nabla \tan^{-1}\left(\frac{y}{x}\right)$$
Now looking at your integral, each term in the summation moves the singularity of the vector field over on the $x$ axis. Meaning we should check when the singularities leave the bounds of the polar curve. Since the singularities are only moving in the $+x$ direction, check the point $\theta=0$, the intersection of the loop with the $+x$ axis:
$$r(0) = 2 + 4\cos(0) = 6$$
therefore the only terms in the summation that contribute to the integral are 
$$\int_\gamma \left(\sum_{n=0}^{10} \frac{-ydx + (x-2n-1)dy}{(x-2n-1)^2+y^2}\right) = \int_\gamma \left(\sum_{n=0}^{2} \frac{-ydx + (x-2n-1)dy}{(x-2n-1)^2+y^2}\right)$$
since the other terms are integrals of conservative vectors fields, hence evaluate to $0$. Now the remaining ones use a curious property of curl-free, nonconservative vector fields, which is that
$$\int_{\gamma_1} \vec{F}\cdot d\vec{r} = \int_{\gamma_2} \vec{F}\cdot d\vec{r}$$
as long as the curves $\gamma_1$ and $\gamma_2$ contain the same singularities. This can be proven using Green's theorem. So for each of the $3$ remaining terms in the summation, we could replace the curve with a unit circle centered at the given singularity instead, resulting in
$$\int_\gamma \left(\sum_{n=0}^{2} \frac{-ydx + (x-2n-1)dy}{(x-2n-1)^2+y^2}\right) = 3 \int_{\text{unit circle}}\frac{-y\:dx+ x\:dy}{x^2+y^2} = 6\pi$$
