determine the integral:
I= $\int_A^B${[$12x^2y+(\frac{y}{x})+4x]$dx + ($2y+4x^3+log_ex$) dy}
where the (x,y) coordinates of A & B are (1,1) and (2,2), respectively.
I tried seperating the integral into two part and then integrated each part with respect to its variable but ended up with a function and not a value , so how can i fully integrate it?
Assuming that you integrate over the straight line between $A$ and $B$:
First, find a parametrization of the curve. Here, $(x,y)=(t,t)$, $1\leq t\leq 2$ will do. Then $dx=dt$ and $dy=dt$, so your integral, according to the definition of curve integral, transforms into $$ \int_1^2 12t^2t+\frac{t}{t}+4t+2t+4t^3+\log_{e} t\,dt. $$ Can you take it from here?
Edit
Since you noted that there exists a potential, and actually had calculated it, here comes a way using it: Let $$ u(x,y)=2x^2+4x^3y+y^2+y\log_e x $$ be the potential. Then, the integral you want to calculate is just $u(2,2)-u(1,1)$, i.e. the value of the potential at the end-point minus the potential at the start-point.