求∫(x^3*根号下x^2-9)的-1次方的不定积分
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令x = 3secθ and dx = 3secθtanθdθ
then cosθ = 3/x and sinθ = √(x² - 9)/x and tanθ = √(x² - 9)/3
∫ 1/[x³√(x² - 9)] dx
= ∫ (3secθtanθ)/[(3secθ)³(3tanθ)] dθ
= ∫ (3secθtanθ)/(81sec³θtanθ) dθ
= (1/27)∫ cos²θ dθ
= (1/54)∫ (1 + cos2θ) dθ
= (1/54)(θ + sinθcosθ) + C
= (1/54)arccos(3/x) + (1/54) · √(x² - 9)/x · 3/x + C
= (1/54)arccos(3/x) + √(x² - 9)/(18x²) + C For x > 3
then cosθ = 3/x and sinθ = √(x² - 9)/x and tanθ = √(x² - 9)/3
∫ 1/[x³√(x² - 9)] dx
= ∫ (3secθtanθ)/[(3secθ)³(3tanθ)] dθ
= ∫ (3secθtanθ)/(81sec³θtanθ) dθ
= (1/27)∫ cos²θ dθ
= (1/54)∫ (1 + cos2θ) dθ
= (1/54)(θ + sinθcosθ) + C
= (1/54)arccos(3/x) + (1/54) · √(x² - 9)/x · 3/x + C
= (1/54)arccos(3/x) + √(x² - 9)/(18x²) + C For x > 3
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