求定积分∫（1到√3）1-x^2/x^2（1 x^2）dx
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∫(1→√3) (1 - x²)/[x²√(1 + x²)] dx
令x = tanθ，dx = sec²θ dθ
x = 1，θ = π/4
x = √3，θ = π/3
= ∫(π/4→π/3) [(1 - tan²θ)(sec²θ)]/(tan²θsecθ) dθ
= ∫(π/4→π/3) (1 - tan²θ) * cosθ/sin²θ dθ
= ∫(π/4→π/3) cscθcotθ dθ - ∫(π/4→π/3) secθ dθ
= - cscθ - ln(secθ + tanθ) |[π/4→π/3]
= - 2/√3 - ln(2 + √3) - [- √2 - ln(1 + √2)]
= √2 - 2/√3 + ln[(1 + √2)/(2 + √3)]
令x = tanθ，dx = sec²θ dθ
x = 1，θ = π/4
x = √3，θ = π/3
= ∫(π/4→π/3) [(1 - tan²θ)(sec²θ)]/(tan²θsecθ) dθ
= ∫(π/4→π/3) (1 - tan²θ) * cosθ/sin²θ dθ
= ∫(π/4→π/3) cscθcotθ dθ - ∫(π/4→π/3) secθ dθ
= - cscθ - ln(secθ + tanθ) |[π/4→π/3]
= - 2/√3 - ln(2 + √3) - [- √2 - ln(1 + √2)]
= √2 - 2/√3 + ln[(1 + √2)/(2 + √3)]
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