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$\int \frac{[\cot (\log x)]^{2}}{x} d x is equal to$

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−cotlogx−logx+C

cotlogx+logx+C

−cotlogx+logx+C

cotlogx−logx+C

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detailed solution

Correct option is A

Let I=∫[cot⁡log⁡x]2xdxlog⁡x=y1xdx=dyI=∫cot2⁡ydy=∫cosec2⁡y−1dy=−cot⁡y−y+C=−cot⁡log⁡x−log⁡x+C

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∫[cot⁡(log⁡x)]2xdx is equal to

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$\int \frac{[\cot (\log x)]^{2}}{x} d x is equal to$