求不定积分（1／x根号x^2-1)dx

---

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

---

arccos 1/x的绝对值+C

---

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'

---