求定积分∫(1-√3)dx/(x√(x^2+1))

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令x=tana
dx=sec²ada
x=√3,a=π/3
x=1,a=π/4
原式=∫(π/4,π/3)sec²ada/(tanaseca)
=∫(π/4,π/3)da/sina
=∫(π/4,π/3)sinada/(1-cos²a)
=-∫(π/4,π/3)dcosa/(1-cosa)(1+cosa)
=-1/2∫(π/4,π/3)[-1/(cosa-1)+1/(1+cosa)]dcosa
=-1/2ln[(1+cosa)/(1-cosa)](π/4,π/3)
=-1/2ln(1+1/2)/(1-1/2)+1/2ln(1+√2/2)/(1-√2/2)
=ln[(√6+√3)/3]

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