Verified answer

First of all, your C is rotated in the wrong direction, so simply replace your solution for its negative. ∫C = -∫-C

The counter-clockwise curve -C can be traced out by the position vector r = < cos t, sin t, 0 >, as t goes from 0 to 2π.

dr/dt = <-sin t, cos t, 0>

dr = <-sin t, cos t, 0> dt

Since we are looking for ∫-C F ⋅ dr we can substitute

∫-C F ⋅ <-sin t, cos t, 0> dt

And all x= r cos t, y= r sin t, z=0, and r=1

F = 1/2<y/z²+1,-x/z²+1,-x²+y²/(z²+1)²> turns into

F = 1/2<sin t, -cos t, -cos² t + sin² t>

∫-C F ⋅ <-sin t, cos t, 0> dt

∫-C 1/2<sin t, -cos t, -cos² t + sin² t> ⋅ <-sin t, cos t, 0> dt

∫-C 1/2 (-sin² t - cos² t + 0) dt

∫-C -1/2 dt

-t/2 ]]-C

-π

But your original question was for the clockwise curve C. So we negative the answer.

π