Going (not quite) in circles

Here's a solution for yah, Sir Otto!

Taking $dx = 1 - 10\sin(t) dt, dy = 10\cos(t) dt$ , we can now transform the above path integral into one for the variable $t:$

$I = \frac{1}{2\pi} \cdot \int_{0}^{10\pi} \frac{(-10\sin(t))(1-10\sin(t))}{(t+10\cos(t))^2 + (10\sin(t))^2} dt + \frac{(t+10\cos(t))(10\cos(t))}{(t+10\cos(t))^2 + (10\sin(t))^2} dt$ ;

or $I = \frac{1}{2\pi} \int_{0}^{10\pi} \frac{100\sin^{2}(t) + 100\cos^{2}(t) - 10\sin(t) + 10t\cos(t)}{100\sin^2(t) + 100\cos^{2}(t) + 20t\cos(t) + t^2} dt;$

or $I = \frac{1}{2\pi} \int_{0}^{10\pi} \frac{100 - 10\sin(t) + 10t\cos(t)}{100 + 20t\cos(t) + t^2} dt;$

or $I = \frac{1}{2\pi} \cdot [\frac{t}{2} + \arctan(\frac{(t+10)\cot(t/2)}{t-10}]|_{0}^{10\pi};$

or $I = \frac{1}{2\pi} \cdot [5\pi + \arctan(\frac{\pi+1}{\pi-1}\cot(5\pi)) - 0 - \arctan(-\cot(0))];$

or $I = \frac{1}{2\pi} \cdot (5\pi - \frac{\pi}{2} - \frac{\pi}{2});$

or $I = \frac{4\pi}{2\pi} = \boxed{2}.$