$ I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt$ is always zero for $r\in[0,1)$. Why? For $r\in [0,1)$ define
$$
I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt.
$$
Numerical experiments hint at $I(r) = 0$ for all $r\in [0,1)$ but I can't show this analytically.
This integral appears when you compute the Cauchy transform of $\overline z$ over the unit circle. The latter therefore seems to be constantly zero.
 A: With $\gamma$ being the counter-clockwise unit circle and $\bar\gamma$ being the clockwise unit circle,
$$
\begin{align}
\int_0^{2\pi}\frac{r-\cos(t)}{1-2r\cos(t)+r^2}\,\mathrm{d}t
&=\int_0^{2\pi}\frac12\left(\frac1{r-e^{it}}+\frac1{r-e^{-it}}\right)\mathrm{d}t\tag1\\
&=\frac1{2i}\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}-\frac1{2i}\int_{\bar\gamma}\frac{\,\mathrm{d}z}{z(r-z)}\tag2\\
&=\frac1i\int_\gamma\frac{\,\mathrm{d}z}{z(r-z)}\tag3\\
&=\frac1{ir}\int_\gamma\left(\frac1z-\frac1{z-r}\right)\mathrm{d}z\tag4\\
&=\left\{\begin{array}{}
0&\text{if }|r|\lt1\\
\frac{2\pi}r&\text{if }|r|\gt1
\end{array}\right.\tag5
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $z=e^{it}$ in the left integral and $z=e^{-it}$ in the right integral
$(3)$: $\bar\gamma$ is in the opposite direction from $\gamma$
$(4)$: partial fractions
$(5)$: $2\pi i$ times the sum of the residues inside $\gamma$
A: Here is yet another slightly different approach.
Let
$$P_r(t) = \frac{\cos t - r}{1 - 2r \cos t + r^2}, \qquad r \in [0,1).$$
Using $\cos t = (e^{it} + e^{-it})/2$, we can rewrite $P_r(t)$ as
\begin{align}
P_r (t) &= \frac{1}{2} \frac{(e^{it} + e^{-it}) - 2r}{1 - r(e^{it} + e^{-it}) + r^2}\\
&= -\frac{1}{2} \frac{2r - e^{it} - e^{-it}}{(r - e^{it})(r - e^{-it})}\\
&= -\frac{1}{2} \left [\frac{1}{r - e^{it}} + \frac{1}{r - e^{-it}} \right ]\\
&= \frac{1}{2} \left [\frac{e^{-it}}{1 - r e^{-it}} + \frac{e^{it}}{1 - r e^{it}} \right ]\\
&= \frac{1}{2} \left [e^{-it} \sum_{n = 0}^\infty r^n e^{-int} + e^{it} \sum_{n = 0}^\infty r^n e^{int} \right ]\\
&= \sum_{n = 0}^\infty r^n \left (\frac{e^{i(n + 1)t} + e^{-i(n + 1)t}}{2} \right )\\
&= \sum_{n = 0}^\infty r^n \cos (n + 1)t.
\end{align}
Now 
\begin{align}
\int^{2\pi}_0 P_r (t) \, dt = \int^{2 \pi}_0 \sum_{n = 0}^\infty r^n \cos (n + 1)t \, dt = \sum_{n = 0}^\infty r^n \int^{2\pi}_0 \cos (n + 1)t \, dt = 0.
\end{align} 
A: Note that $$I(r) =\frac{-1}{2r}\int_{0}^{2\pi}\frac{1-2r\cos t+r^2+r^2-1}{1-2r\cos t+r^2}\,dt$$ which is same as $$-\frac{\pi} {r} +\frac{1-r^2}{r}\int_{0}^{\pi}\frac{dt}{1+r^2-2r\cos t} $$ The integral above is equal to $$\frac{\pi} {\sqrt{(1+r^2)^2-4r^2}}=\frac{\pi}{1-r^2} $$ and hence $I(r) =0$.

We have used the formula $$\int_{0}^{\pi}\frac{dt}{a+b\cos t} =\frac{\pi} {\sqrt{a^2-b^2}},\,a>|b|$$ which can be proved using the substitution $$(a+b\cos t) (a-b\cos u) =a^2-b^2$$
A: Hint: $$\cos(t)=\frac{1-a^2}{1+a^2}$$ and $$dt=\frac{2da}{1+a^2}$$
Now your indefinite integral is given by $$\int-\frac{2 \left(a^2 r+a^2+r-1\right)}{\left(a^2+1\right)
   \left(a^2 r^2+2 a^2 r+a^2+r^2-2 r+1\right)}da$$ and this is rational
The solution of this integral is given by $$-2 \left(\frac{\tan ^{-1}(a)}{2 r}-\frac{\tan
   ^{-1}\left(\frac{a (r+1)}{1-r}\right)}{4
   r}+\frac{\tan ^{-1}\left(\frac{a
   (r+1)}{r-1}\right)}{4 r}\right)$$
A: Not sure why none of the answers mention this approach, but you can just take the derivative with respect to $r$:
\begin{align}
I'(r) &= \int_0^{2\pi}\frac{d}{dr}\left(\frac{\cos t - r}{1 - 2r\cos t + r^2}\right)\,dt\\
&= \int_0^{2\pi} \frac{(r^2-1)-2r\cos t + 2\cos^2t}{(1 - 2r\cos t + r^2)^2} \\
&= \frac1{i}\int_{|z|=1} \frac{(r^2-1)-r(z+z^{-1}) + \frac12(z+z^{-1})^2}{(1 - r(z+z^{-1}) + r^2)^2}\cdot\frac{dz}{z}\\
&= \frac1{i}\int_{|z|=1} \frac{z^4-2r z^3+2r^2z^2-2rz+1}{(r^2+1)z(z-r)^2\left(z-\frac1r\right)^2}\,dz\\
\end{align}
Since $r \in [0,1\rangle$, the relevant residues are $$\operatorname{Res}(f,0) = \frac1{2r^2}$$
$$\operatorname{Res}(f,r) = -\frac1{2r^2}$$
so $I'(r) = 0$ for $r \in [0,1\rangle$.
We conclude that $I$ is constant on $[0,1\rangle$. Hence $$I(r)=I(0) = \int_0^{2\pi}\cos t\,dt = 0$$
A: An alternative proof using complex methods:
For $0< r < 1$ let
$$ f_r \colon B_\frac{1}{r} (0) \to \mathbb{C} \, , \, f_r(z) = \frac{- \ln(1-rz)}{z} \, , $$
where $f_r(0) = r $ . Then $f_r$ is holomorphic, so
$$ I(r) \equiv - \int \limits_0^{2\pi} \ln(1-r \mathrm{e}^{\mathrm{i}t}) \, \mathrm{d} t = - \mathrm{i} \int \limits_{S^1} f_r(z) \, \mathrm{d} z = 0 $$
holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm:
$$ I(r) = \sum \limits_{n=1}^\infty \frac{r^n}{n} \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} n t} \, \mathrm{d} t = 0 \, . $$
This implies
\begin{align}
\int \limits_0^{2\pi} \frac{\cos(t) - r}{1 - 2 r \cos(t) + r^2} \, \mathrm{d} t &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln(1 - 2 r \cos(t) + r^2) \, \mathrm{d} t \\
&= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln[(1 -r \mathrm{e}^{\mathrm{i} t})(1 -r \mathrm{e}^{-\mathrm{i} t})] \, \mathrm{d} t \\
&= \frac{\mathrm{d}}{\mathrm{d} r} I(r) = \frac{\mathrm{d}}{\mathrm{d} r} 0 = 0
\end{align}
as desired.
A: (Since it is not mentioned yet) The function
$$
P_r(t) = \sum_{n=-\infty}^\infty r^{|n|} e^{int} =\frac{1-r^2}{1-2r\cos t +t^2},\quad0\le r<1, 0\le t\le 2\pi
$$ is called the Poisson kernel. For any continuous function $f:[0,2\pi]\to \Bbb C$ the Poisson integral
$$
u(r,\theta) = \frac{1}{2\pi i}\int_0^{2\pi} f(t)P_r(\theta-t)\mathrm{d}t\tag{*}
$$ gives the unique solution of the Dirichlet problem $\triangle u =0, \lim\limits_{r\to 1^-}u(r,\theta)=f(\theta)$.
Let $u(r,\theta)=r\cos \theta$. Since it is a real part of the analytic function $z=r e^{i\theta}$, we find that $$\triangle u=0,\ \ \ \lim\limits_{r\to 1^-}u(r,\theta)=\cos \theta.$$ By $\text{(*)}$ it follows that
$$
r\cos \theta =\frac{1}{2\pi}\int_0^{2\pi} \frac{\cos t(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t
$$ for all $0\le r<1$ and $0\le \theta\le 2\pi$. Using the fact
$$
\frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t=\frac{1}{2\pi}\sum_{n=-\infty}^\infty r^{|n|} \int_0^{2\pi}e^{in(t-\theta)}\mathrm{d}t =1,
$$ it follows
$$
\frac{1}{2\pi}\int_0^{2\pi} \frac{\left(\cos t-r\cos \theta\right)(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t =0.
$$ By letting $\theta =0$, we obtain
$$
\frac{1-r^2}{2\pi }\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0,$$
or equivalently
$$
\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0.$$
A: It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.
Here, consider the integral
$$
\vec I(M) = \int_{X\in C}\frac{\vec{MX}}{MX^2}\mathrm ds
$$
where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $\mathrm ds$ is a length element of the circle. What you are asking for is a component of $\vec I$, the one directed along $\vec{OM}$. In fact, we have $\vec I=\vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.
A: From the identity
$$\sum_{n=1}^{\infty}r^{n-1} \cos(nt)=\frac{\cos(t)-r}{1-2r\cos(t)+r^2}, \ |r|<1$$
it follows that 
$$I(r)=\sum_{n=1}^\infty r^{n-1}\int_0^{2\pi}\cos(nt)dt=\sum_{n=1}^\infty\frac{r^{n-1}\sin(2n\pi)}{n}=0$$
