_ ^ [ 1 x - 1 ( x) ^2 ]dx = - Vidyadip

MCQ

$\int_{}^{} {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]dx = } $

A
$\frac{1}{{\log x}} + c$
✓
$\frac{x}{{\log x}} + c$
C
$\frac{x}{{{{(\log x)}^2}}}$
D
None of these

✓

Answer

Correct option: B.

$\frac{x}{{\log x}} + c$

b
(b)$\int_{}^{} {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx = \int_{}^{} {\frac{1}{{\log x}}\,dx - \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}\,dx} } $
$ = \frac{x}{{\log x}} + \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}\,.\,\frac{1}{x}x\,dx} - \int_{}^{} {\frac{1}{{{{(\log x)}^2}}}} dx + c = \frac{x}{{\log x}} + c$.

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