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The problem is
1) Show that the line integral $\displaystyle \int ye^{xy}dx+ xe^{xy}dy$ is independent of the path.
2) Find the integral from (1, 0) to (0, 4)
A line integral such as that is independent of the path if and only if the differential being integrated is an "exact differential"- that is, that there exist a function F(x,y) such that $\displaystyle dF= F_xdx+ F_ydy= ye^{xy}dx+ (xe^{xy}- 2y)dy$. Since we must have [tex]F_{xy}= F_{yx}, that, in turn, is true if and only if $\displaystyle (ye^{xy})_y= (xe^{xy}- 2y)_x$. Is that true?
Once you have shown that, you can do (2) by actually finding that F(x,y).
We must have $\displaystyle F_x= ye^{xy}$. Integrating that, treating y as a constant, $\displaystyle F(x,y)= e^{xy}+ g(y)$. (Since we are treating y as a constant, the "constant of integration" may be a function of y- that is the "g(y)".)
From $\displaystyle F(x,y)= e^{xy}+ g(y)$ and so $\displaystyle F_y= xe^{xy}+ g'(y)= ye^{xy}-2y$. The "$\displaystyle ye^{xy}$" cancels out- that had to happen in order that g' be a function of of y only and happens precisely because of the condition in (1). That tells us g'(y)= -2y so that $\displaystyle g(y)= -y^2+ C$ where C now really is a constant. That is, $\displaystyle F(x,y)= e^{xy}- 2y$. Evaluate that at (0,4) and (1, 4) and subtract.