求三重积分∫∫∫1/(x+y+z)^2,Ω：0

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求三重积分∫∫∫1/(x+y+z)^2,Ω：0
突然发现题弄错了，是3次方。求三重积分∫∫∫1/(x+y+z)^3,Ω：0

∫∫∫1/(x+y+z)²dxdydz
=∫∫[0--->1]∫[0--->1] 1/(x+y+z)² dxdydz
=∫[0--->1]∫[0--->1] -1/(x+y+z) |[1--->2]dydx
=∫[0--->1]∫[0--->1] [1/(x+y+1)-1/(x+y+2)]dydx
=∫[0--->1] [ln(x+y+1)-ln(x+y+2)] |[0--->1]dx
=∫[0--->1] [ln(x+1+1)-ln(x+0+1)-ln(x+1+2)+ln(x+0+2)] dx
=∫[0--->1] [2ln(x+2)-ln(x+1)-ln(x+3)] dx
分部积分
=x[2ln(x+2)-ln(x+1)-ln(x+3)]-∫[0--->1] [2x/(x+2)-x/(x+1)-x/(x+3)]dx
=x[2ln(x+2)-ln(x+1)-ln(x+3)]-∫[0--->1] [2(x+2-2)/(x+2)-(x+1-1)/(x+1)-(x+3-3)/(x+3)]dx
=x[2ln(x+2)-ln(x+1)-ln(x+3)]-∫[0--->1] [2-4/(x+2)-1+1/(x+1)-1+3/(x+3)]dx
=x[2ln(x+2)-ln(x+1)-ln(x+3)]-∫[0--->1] [-4/(x+2)+1/(x+1)+3/(x+3)]dx
=x[2ln(x+2)-ln(x+1)-ln(x+3)]+ [4ln(x+2)-ln(x+1)-3ln(x+3)] |[0--->1]
=2ln3-ln2-ln4+4ln3-ln2-3ln4-4ln2+3ln3
=9ln3-6ln2-4ln4
=9ln3-6ln2-8ln2
=9ln3-14ln2
再问：额，，突然发现题弄错了，是3次方。。。求三重积分∫∫∫1/(x+y+z)^3,Ω：02] dydx =(1/2)∫[0--->1]∫[0--->1] [1/(x+y+1)²-1/(x+y+2)²] dydx =(1/2)∫[0--->1] [-1/(x+y+1)+1/(x+y+2)] |[0--->1]dx =(1/2)∫[0--->1] [-1/(x+1+1)+1/(x+0+1)+1/(x+1+2)-1/(x+0+2)] dx =(1/2)∫[0--->1] [1/(x+1)+1/(x+3)-2/(x+2)] dx =(1/2)[ln(x+1)+ln(x+3)-2ln(x+2)] |[0--->1] =(1/2)[ln2+ln4-2ln3-ln3+2ln2] =(1/2)[3ln2+ln4-3ln3] =(1/2)[3ln2+2ln2-3ln3] =(1/2)(5ln2-3ln3)

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