(1+x)dx-(1-y)dy=(dx-dy)+(xdx+ydy)=d(x-y)+d(x^2+y^2)=0
即d(x-y)=-d(x^2+y^2)
两端积分,得x-y=-(x^2+y^2)/2+c
所以,(x^2+y^2)/2+x-y=c