答:
∫(0到π/2)dθ∫(0到1)ln(1+r^2)rdr
算不定积分∫rln(1+r^2)dr
=∫1/2ln(1+r^2)d(1+r^2)
=1/2∫ln(1+r^2)d(1+r^2)
∫lnxdx=xlnx-x+C
所以1/2∫ln(1+r^2)d(1+r^2)
=1/2[(1+r^2)ln(1+r^2)-(1+r^2)]+C
则∫(0到π/2)dθ∫(0到1)ln(1+r^2)rdr
=π/2∫(0到1)ln(1+r^2)rdr
=π/2[1/2((1+r^2)ln(1+r^2)-(1+r^2))]|(0到1)
=π/4(2ln2-2-(-1))
=(2ln2-1)π/4