Consider the vector field:
$$F=\left(\frac{2xy-2xy^2}{\left(1+x^2\right)^2}+\frac{8}{13}\right)i+\left(\frac{2y-1}{1+x^2}+2y\right)j$$
Determine $$\int_C F\cdot dr$$ where $C$ is the path $C_1+C_2+C_3$ from $(2,0)$ to $(5,6)$ shown.
I already found out that F is path-independent by the Curl test, but I'm not sure what to do from here? I don't have to parameterize each one of these curves, do I? That seems excessive and there should be a simple way to solve this.
$$\frac{\partial f}{\partial x}=\frac{2x(y-y^2)}{(1+x^2)^2}+\frac{8}{13}$$ $$\frac{\partial f}{\partial y}=\frac{2y-1}{1+x^2}+2y$$ Therefore $$f=\frac{y^2-y}{1+x^2}+\frac{8x}{13}+y^2$$ You can solve for the gradient function by integrating each equation with respect to its appropriate differential, then solve for the arguments which are exclusively of the other variables. By the fundamental theorem of line integrals, $f(5,6)-f(2,0)=\int_{C} F \cdot dr$.