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

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$cotlog x + \log x + C$

$- cotlog x + \log x + C$

$cotlog x - \log x + C$

$- cotlog x - \log x + C$

answer is A.

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Detailed Solution

$Let I = \int \frac{[\cot \log x]^{2}}{x} d x$

$\begin{array}{l} \log x = y \\ \frac{1}{x} d x = d y \\ I = \int \cot^{2} y d y \\ = \int ({cosec}^{2} y - 1) d y \\ = - \cot y - y + C \\ = - \cot \log x - \log x + C \end{array}$

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