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rusak2 [61]
3 years ago
8

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

Mathematics
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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\boxed{\bold{\huge{\boxed{497}}}}

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