一步一步的积分即可∫(0~1-x-y)1/(1+x+y+z)^3dz = [-1/2(1+x+y+z)^2] (0~1-x-y) = 1/2(1+x+y)^2 - 1/8∫ ( 1/2(1+x+y)^2 - 1/8 ) dy = 1/2(1+x) - (3-x) / 8 积分限(0,1-x)∫ ( 1/2(1+x) - (3-x) / 8 ) dx = ln(1+x)...