第1个回答 2012-03-12
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本回答被提问者采纳
∫(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本回答被提问者采纳
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