∫∫D[(1-x2-y2)/(1+x2+y2)]dσ，其中D是x2+y2=1，x=0,y=0所围第I象限内的部分．

∫∫_D[(1-x²-y²)/(1+x²+y²)]dσ，其中D是
x²+y²=1，x=0,y=0所围第I象限内的部分．
【正确答案】：∫∫_D[(1-x²-y²)/(1+x²+y²)]dσ= ∫₀^π/2dθ∫₀¹[(1-r²)/(1+r²)]rdr= π/2∫₀¹[2/(1+r²)-1]rdr =π/2[ln(1+r²)∣₀¹-(1/2)r²∣₀¹] =π/2(ln2-1/2)