求解不定积分∫dx/√(9x^2-6x+7)

Question

求解不定积分∫dx/√(9x^2-6x+7)

Answer 1

I = ∫ 1/√(9x² - 6x + 7) dx
= ∫ 1/√[(3x - 1)² + 6] dx
令3x - 1 = √6tanθ,3 dx = √6sec²θ dθ
I = ∫ 1/√(6tan²θ + 6) * (√6/3)sec²θ dθ
= ∫ 1/(√6secθ) * (√6/3)sec²θ dθ
= (1/3)∫ secθ dθ
= (1/3)ln|secθ + tanθ| + C
= (1/3)ln|tanθ + √(1 + tan²θ)| + C
= (1/3)ln|(3x - 1)/√6 + √[1 + (3x - 1)²/6]| + C
= (1/3)ln|3x - 1 + √(9x² - 6x + 7)| + C
佩服楼上了,∫ √(9x² - 6x + 7) dx的做法不见得比∫ 1/√(9x² - 6x + 7) dx容易,而且会复杂得多.