I = ...... = ∫<3/4, 4/3>√(1+θ^2)dθ/θ^2令 θ = tanu， 则I = ∫(secu)^3du/(tanu)^2= ∫du/[cosu(sinu)^2]= ∫dsinu/[(cosu)^2 (sinu)^2]= ∫dsinu/{[1-(sinu)^2](sinu)^2}= ∫[1/(sinu)^2 + (1/2)[1/(1-sinu) + 1/(1+sinu)]dsinu= [-cotu +(1/2)ln{(1+sinu)/(1-sinu)}]= 7/12 + ln(3/2)
