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B
14log(ab)⋅logba
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C
log(ab)⋅logba
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D
−12log(ab)⋅logba
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Solution
The correct option is A12log(ab)⋅logba Substitute logx=t∴(1/x)dx=dt and adjust the limits ∴I=∫tdt=[12t2]logbloga ∴I=12[(logb)2−(loga)2] =12[(logb+loga)(logb−loga)] ∴I=12log(ab)⋅logba.