Seeking indefinite integral ∫ (X & # 178; - 9) ^ 1 / 2 / xdx

Seeking indefinite integral ∫ (X & # 178; - 9) ^ 1 / 2 / xdx

Let x = 3sec θ, DX = 3sec θ Tan θ D θ, √ (X & # 178; - 9) = √ (9sec & # 178; θ - 9) = 3tan θ, 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

Elden Ring Premiere

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