Question

根号下x^2-1/x的不定积分

Answer 1

根号下X^2-1-ACOS(1/X)

Answer 2

令x=secy，dx=secytany dy
∫ √(x²-1) / x dx
= ∫ tany / secy * secytany dy
= ∫ tan²y dy
= ∫ (sec²y-1) dy
= tany - y + C
= √(x²-1) - arcsec(x) + C