Integrate x2−√xx with respect to x
x22−2√x+K
2(x2−x)3x+K
x22−√x+K
(x2−x)3x+K
Correct answer is A
∫x2−√xx=∫x2x−x12x∫x−x−12=(12)x2−x1212+K=x22−2x12+K=x22−2√x+K
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