Question

如图所示，求详解过程

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Answer 1

令 u(x,y) =∫[(y2+2xy-x^2)/(x^2+y^2)^2] dx
=2y^2*∫[1/(x^2+y^2)^2] dx + y∫[1/(x^2+y^2)^2] d(x^2) -∫[1/(x^2+y^2)] dx
= (1/y)[xy/(x^2+y^2) + arctan(x/y)] - y/(x^2+y^2) - (1/y)*arctan(x/y) + h(y
= (x-y)/(x^2+y^2) + h(y)
由 du/dy = (y2-2xy-x^2)/(x^2+y^2)^2 +h'(y) = (y2-2xy-x^2)/(x^2+y^2)^2,
得到 h'(y)=0 , 即 h(y)=C
所以， u(x,y)= (x-y)/(x^2+y^2) + C

Answer 2

u(x,y) = ∫<下0,上x>P(x,0)dx + ∫<下0,上y>Q(x,y)dy
= ∫<下0,上x>(-1/x^2)dx - ∫<下0,上y>[(x^2+2xy-y^2)/(x^2+y^2)^2]dy
= 1/x - ∫<下0,上y>[(x^2+2xy-y^2)/(x^2+y^2)^2]dy ,
令 y=xtant， 则 dy=x(sect)^2dt,
I = ∫[(x^2+2xy-y^2)/(x^2+y^2)^2]dy
= (1/x)∫{[1+2tant-(tant)^2]/(sect)^2}dt
= (1/x)∫(cost)^2[1+2tant-(tant)^2]dt
= (1/x)∫(cos2t+sin2t)dt
= (1/x)(1/2)[sin2t-cos2t]
因 tant=y/x, 得 sin2t=2xy/(x^2+y^2), cos2t=(x^2-y^2)/(x^2+y^2),
则 I = [1/(2x)](y^2+2xy-x^2)/(x^2+y^2)
u(x,y) = 1/x-[1/(2x)][(y^2+2xy-x^2)/(x^2+y^2)]<下y=0, 上y=y>
= 1/x-[1/(2x)][(y^2+2xy-x^2)/(x^2+y^2)]<下y=0, 上y=y>
= [1/(2x)][1-(y^2+2xy-x^2)/(x^2+y^2)]
= (x-y)/(x^2+y^2)

追问

P和Q在（0,0）处无定义，不能从（0,0）开始积吧

追答

从（0+,0+）开始积, 结果一样。