求不定积分（1／x根号x^2-1)dx

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令x = secz，dx = secztanz dz
∫ 1/[x√(x²薯竖 - 1)] dx
= ∫ 1/[secz · √友旅(sec²z - 1)] · secztanz dz
= ∫ 1/好手凳[secz · tanz] · secztanz dz
= ∫ dz
= z + C
= arcsec(x) + C
= arccos(1/x) + C

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arccos 1/x的绝对值+C

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S（1／x根号x^2-1)dx=S（1／x根号x^2)dx-S(1)dx=ln|x|x||+C-x-C'

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