find the integral of 1/sqrt(x^2-9)
Let x = 3secβ, then dx = 3secβtanβ∫dx/√(x²-9) = ∫[(3secβtanβ)/(3tanβ)]dβ = ∫secβdβ = ln |tanβ - secβ| + c = ln |√(x² - 9)/3 - x/3| + C = ln |√(x² - 9)/x| + C
