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∫{1log⁡ex−1(log⁡ex)2}dx\int\left\{\dfrac{1}{\log_e x} - \dfrac{1}{(\log_e x)^2}\right\}dx is equal to : (where CC is an arbitrary constant)

Solution

✅ Correct Option: 2

Observe that ddx(xlog⁡ex)=log⁡ex−1(log⁡ex)2=1log⁡ex−1(log⁡ex)2\dfrac{d}{dx}\left(\dfrac{x}{\log_e x}\right)=\dfrac{\log_e x-1}{(\log_e x)^2}=\dfrac{1}{\log_e x}-\dfrac{1}{(\log_e x)^2}, exactly the integrand. Hence the integral is xlog⁡ex+C\dfrac{x}{\log_e x}+C.

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