1 log x e ? d x =

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Solution

Q.
$\int \frac{1}{\log_{x} e} d x =$

1 $\log (\log_{x} e) + C$

2 $\frac{1}{{(\log_{x} e)}^{2}} + C$

3 $x \log_{e} (\frac{x}{e}) + C$

4 None of these

Ans.
(3)

Ans. $x \log_{e} (\frac{x}{e}) + C$

Explanation:

$\int \frac{1}{\log_{x} e} d x = \int \log_{e} x d x$

$= \int 1 \cdot \log_{e} x d x$

   $= x \log_{e} x - \int \frac{1}{x} \cdot x d x$

      $= x \log_{e} x - x + C$

$= x (\log_{e} x - \log_{e} e) + C$

    $= x \log_{e} (\frac{x}{e}) + C$

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