Confusion in integrating multivariable function 
Integrate $\int_{C}{\frac{-x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy}$
  C: $x=cost$, $y=sint$, $\quad0\le t\le \frac{\pi}{2}$


In this case, It's incorrect to integrate it as $\frac{-1}{2}\ln{(x^2+y^2)}|_{a}^{b}+\frac{1}{2}\ln{(x^2+y^2)}|_{c}^{d}$.
but,


$\int_{0}^{1}\int_{0}^{1}{\frac{1}{1-xy}dxdy}=\int_{0}^{1}\int_{0}^{1}{\frac{-1}{y}\frac{-y}{1-xy}dxdy}=\int_{0}^{1}{\frac{-1}{y}[\ln({1-xy}]_{0}^{1})}dy$


I believe this work.  
My question is when integrands are multivariable functions for both cases, why does only the bottom case work?
Is it about integration boundary? why is it exactly?

 A: It is useful to understand the intuition behind in order to understand the distinction. 
The line integral gives you the amount of work that the vector field
$$
F(x,y) = \left( \frac{-x}{x^2 + y^2}, \frac{y}{x^2 + y^2} \right),
$$
does on a particle that moves from $(1,0)$ to $(0,1)$ counterclockwise along the circle that is paremetrized by $r(t) = ( \cos t, \sin t)$.
While the second integral gives you the volume that lies below $f(x,y) = \frac{1}{1 - xy}$ and above the $xy$ plane in the $[0,1]\times[0,1]$ square.
For the line integral you have to account for the vector field, i.e., a "force" that this field apply on a particle along the curve, while for the double integral no "force" is considered and the integration is over a region and not a curve.  
A: Sorry for writing this as an answer, I don't reach 50 (reputation) yet.
You have to understand that in the first case you are supposed to integrate on a curve but in the second you've been given a square: $\,[0,1]\times[0,1]\,$. It is completely different.
