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The integrand 1/(x² + 9) leads to which inverse tangent form?
A(1/9) tan⁻¹(x/9) + C
Btan⁻¹(x/3) + C
C3 tan⁻¹(3x) + C
D(1/3) tan⁻¹(x/3) + C
Answer & Solution
Correct answer: D. (1/3) tan⁻¹(x/3) + C
Using the form ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) with a = 3 gives (1/3)tan⁻¹(x/3).
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