∫∫(ydxdy )/(1+x∧2+y∧2)∧3/2,0

原式=∫(0,1)dx∫(0,1)ydy/√(1+x²+y²)³
=(1/2)∫(0,1)dx∫(0,1)d(1+x²+y²)/√(1+x²+y²)³
=(1/2)∫(0,1){[(-2)/√(1+x²+y²)]│(0,1)}dx
=∫(0,1)[1/√(1+x²)-1/√(2+x²)]dx
={ln[x+√(1+x²)]-ln[x+√(2+x²)]}│(0,1)
=ln(1+√2)-ln(1+√3)-ln1+ln(√2)
=ln[√2(1+√2)/(1+√3)]
=ln[(2+√2)/(1+√3)].
注：你的答案与我的答案有点出入,我觉得是打错了.