对dz/dx做x的积分,有∫ [(x^2+y^2)/x]dx=∫ xdx+y²∫ 1/xdx=x²/2+y²lnx+f(y)代入x=1,z=siny,得z(1,y)=0.5+f(y)=siny所以f(y)=siny-0.5所以z(x,y)=x²/2+y²lnx+siny-0.5