1.{1/y[1+y^2f(xy)]}'y=(-1/y^2)+f(xy)+xyf'(xy)
{x/y^2[y^2f(xy)-1]}'x=(-1/y^2)+f(xy)+xyf'(xy)
故曲线积分I与路径无关.
2.ab=cd
选取(a,b)到(c,b)再到(c,d)
I=∫(a,c)1/b[1+b^2f(xb)]dx+∫(b,d)c[f(cy)-1/y^2]dy
=(a-c)/b+∫(a,c)f(xb)]dbx+∫(b,d)cf(cy)dcy-1/d+1/b
=(a-c)/b-1/d+1/b+F(bc)-F(ba)+F(dc)-F(bc) (F是f的原函数）
=a/b-a/d-1/d+1/b
=(a+1)(1/b-1/d)