Find the integral (∫) of the function 16x3 - 6x2 + 4x -3
A) ∫ (16x3 - 6x2 + 4x -3) dx = 16x4 + 2x3 + 2x2 + 3x + C
B) ∫ (16x3 - 6x2 + 4x -3) dx = 16x4 - 2x3 + 2x2 - 3x + C
C) ∫ (16x3 - 6x2 + 4x -3) dx = 4x4 + 2x3 + 2x2 + 3x + C
D) ∫ (16x3 - 6x2 + 4x -3) dx = 4x4 - 2x3 + 2x2 - 3x + C
Correct Answer:
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