求不定积分∫（x²-9）^1/2/xdx

题目

求不定积分∫（x²-9）^1/2/xdx

答案

令x = 3secθ,dx = 3secθtanθ dθ,√(x² - 9) = √(9sec²θ - 9) = 3tanθ,x > 3
∫ √(x² - 9)/x dx
= ∫ √(9sec²θ - 9)/(3secθ) · (3secθtanθ dθ)
= ∫ 3tanθ · tanθ dθ
= 3∫ sec²θ - 1 dθ
= 3tanθ - 3θ + C
= 3 · √(x² - 9)/3 - 3arcsec(x/3) + C
= √(x² - 9) - 3arccos(3/x) + C