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in Integrals calculus by (36.9k points)

For any natural number m, evaluate ∫ (x3m + x2m + xm)(2x2m + 3xm + 6)1/m dx.

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Best answer

∫ (x3m + x2m + xm)(2x2m + 3xm + 6)1/m dx 

= ∫ (x3m– 1 + x2m– 1 + xm– 1) × x × (2x2m + 3xm + 6)1/m dx = ∫ (x3m– 1 + x2m– 1 + xm–1 )(2x3m + 3x2m + 6xm)1/m dx 

Let 2x3m + 3x2m + 6xm = t

6m(x3m– 1 + x2m– 1 + xm–1) dx = dt

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