To evaluate the integral 2х — 1 dx x2 – 16 we use the partial fraction decomposition of the integrand. Select one: 2х-1 A В x2-16 х-4 x²+4 2х-1 A В x-4 + x+4 x2 -16 х-4 O None of them. 2x-1 A В x2 -16 x2 -4 x+4 2х-1 Ax+B ||

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A: The integration of the integral is√(16-x2).

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