_ ^ 1 _x e dx =

MCQ

$\int_{}^{} {\frac{1}{{{{\log }_x}e}}dx = } $

A
$\log {\log _x}e + c$
B
$\frac{1}{{{{({{\log }_x}e)}^2}}} + c$
✓
$x\log \left( {\frac{x}{e}} \right) + c$
D
None of these

✓

Answer

Correct option: C.

$x\log \left( {\frac{x}{e}} \right) + c$

c
(c)$\int_{}^{} {\frac{1}{{{{\log }_x}e}}\,dx = \int_{}^{} {{{\log }_e}x\,dx} = x\log x - x + c} $
$ = x({\log _e}x - {\log _e}e) + c = x{\log _e}\left( {\frac{x}{e}} \right) + c.$

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