Using integration by substitution evaluate fxV4x² + 5 dx. 12 V (4x2 + 5)4 +C 12 V (4x2 + 5)4 +C 1 V(4x² + 5)5 + C 40 1 V(4x? + 5)5 +C 10 5 V(4x² + 5)5 + C 14 OV(4x2 + 5)4 + C 8

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