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 Evaluate dx(2x7)(x3)(x4)

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a
sec−1⁡(2x−7)+c
b
12sec−1⁡(2x−7)+c
c
17sec−1⁡(2x−7)+c
d
None of these

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detailed solution

Correct option is B

I=∫dx(2x−7)(x−3)(x−4)=∫dx(2x−7)x2−7x+12=12∫2dx(2x−7)(2x−7)2−1=12sec−1⁡(2x−7)+c      ∵ ∫dttt2-1=sec-1(t)+c


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