∫∫x²/(1+y²)dxdy 0<=x<=1,0<=y<=1
=∫1/(1+y²) [x^3/3] dy 0<=x<=1,0<=y<=1
=(1/3)∫1/(1+y²)dy 0<=y<=1
let y = tana
dy =(seca)^2da
y=0, a=0
y=1, a=π/4
(1/3)∫1/(1+y²)dy 0<=y<=1
=(1/3)∫da 0<=a<=π/4
= π/12
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