Calculating the work integral of a vector field. Let F be the vector field 
F = $<\dfrac{1}{x} + 2x + y, \ \ \dfrac{1}{y} + x + 1>$ .
Compute the work integral $\int_{c} F dr$ where C is the path
r(t) = (1 + cos t)i + (2 + sin t) j $ \ \ \ \ -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}$.
I am very confused as to how to approach this problem. I understand that we simply have to parameterize F, find v(t), and dot product both of them, but  given these expressions that would be very tedious. Is there a simpler way to approach this problem?
 A: Fundamental Theorem for Line Integrals
$$\eqalign{
  & W = \int\limits_C {F \bullet dr}   \cr 
  &  -  -  -  - Test\;For\;Gradient\;Field  \cr 
  & F = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {M\;,\;N} \right)  \cr 
  & F = \nabla f\;\;if\;\;\;{\partial _y}M = {\partial _x}N  \cr 
  & {\partial _y}M = 1  \cr 
  & {\partial _x}N = 1  \cr 
  & W = \int\limits_C {F \bullet dr}  = \int\limits_C {\nabla f \bullet dr}  = f({p_1}) - f({p_0})  \cr 
  & \nabla f = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {{\partial _x}f\;,\;{\partial _y}f} \right)  \cr 
  &  -  -  -  - Find\;\;f(x,y)  \cr 
  & \left\{ \matrix{
  {\partial _x}f = {1 \over x} + 2x + y \hfill \cr 
  {\partial _y}f = {1 \over y} + x + 1 \hfill \cr}  \right.  \cr 
  & \int {{\partial _x}fdx = \ln (x) + {x^2}}  + yx + g(y) = f(x,y)  \cr 
  & {\partial _y}f = x + g'(y)  \cr 
  & g'(y) = {1 \over y} + 1  \cr 
  & g(y) = \ln (y) + y + c  \cr 
  & f(x,y) = \ln (x) + {x^2} + yx + \ln (y) + y + c  \cr 
  &  -  -  -  - Find\;\;{p_1}\;and\;{p_0}  \cr 
  & x = 1 + \cos t  \cr 
  & y = 2 + \sin t  \cr 
  & {{ - \pi } \over 2} \le t \le {\pi  \over 2}  \cr 
  & 1 \le x \le 1  \cr 
  & 1 \le y \le 3  \cr 
  & {p_0} = (1,1)  \cr 
  & {p_1} = (1,3)  \cr 
  &  -  -  -  - Finding\;The\;Work  \cr 
  & W = \int\limits_C {F \bullet dr}  = \int\limits_C {\nabla f \bullet dr}  = f({p_1}) - f({p_0})  \cr 
  & W = \left( {\ln (1) + {1^2} + 3 + \ln (3) + 3} \right) - \left( {\ln (1) + {1^2} + 1 + \ln (1) + 1} \right) = \ln (3) + 4 \cr} $$
