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|x|dx is equal to 

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a
12x2+C
b
−x22+C
c
x|x|+C
d
12x|x|+C

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detailed solution

Correct option is D

Let I=∫|x|⋅1dx=|x|x−∫|x|xxdx=x|x|−∫|x|dx⇒2I=x|x|⇒I=x|x|2+C


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