令x = tanθ,dx = sec²θdθ,x∈[1,√3]→θ∈[π/4,π/3]∫(1~√3) 1/[x²√(1 + x²)] dx= ∫(π/4~π/3) sec²θ/(tan²θsecθ) dθ= ∫(π/4~π/3) 1/cosθ • cos²θ/sin²θ dθ= ∫(π/4~π/3) cscθcotθ dθ= - cscθ |(π/4~π/3)= - 1/sin(π/3) + 1/sin(π/4)= √2 - 2/√3