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Q.

∫ex(2 ex−3)2dx =

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a

12 (2ex−3)+C

b

−12 (2ex−3)+C

c

−1(2ex−3)+C

d

12 (2ex+3)+C

answer is B.

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Detailed Solution

2ex−3=t 2exdx=dt 12∫1t2dt=12−1t+C =−121(2ex−3)+C

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∫ex(2 ex−3)2dx =