Evaluate ∫2(2x−3)23dx
...Evaluate ∫2(2x−3)23dx
3/5(2x-3)5/3 + k
6/5(2x-3)5/3 + k
2x-3+k
2(2x-3)+k
Correct answer is A
∫2(2x−3)23dx
Let u=2x−3
du=2dx
= ∫u23du
= u5353+k
= 35(2x−3)53+k
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