−2x−2−73x3+52x2−6x
2x2+73x3−5x+6
12x2+14x−5
−12x−4−14x+5
Correct answer is A
∫(4x−3−7x2+5x−6)dx
= 4x−3+1−3+1−7x2+12+1+5x1+11+1−6x
= −2x−2−73x3+52x2−6x
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