du(x,y)=2xycos(x^2y)dx+x^2cos(x^y)dy,求u(x,y)

因为du/dx=2xycos(x^2y)所以u(x,y)=∫2xycos(x^2y)dx=∫cos(x^2y)d(x^2y)=sin(x^2y)+A(y) 其中A(y)是关于y的任意函数因为du/dy=x^2cos(x^2y)所以x^2cos(x^2y)+A'(y)=x^2cos(x^2y)所以A'(y)=0A(y)=C所以u(x,y)=sin(x^...