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3 years ago

8

Find the integral of 1/sqrt(x^2-9)

1 answer:

Vesna [10]3 years ago

6 0

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

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