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

MCQ

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

A
$x\log (\log x) + \frac{x}{{\log x}} + c$
✓
$x\log (\log x) - \frac{x}{{\log x}} + c$
C
$x\log (\log x) + \frac{{\log x}}{x} + c$
D
$x\log (\log x) - \frac{{\log x}}{x} + c$

✓

Answer

Correct option: B.

$x\log (\log x) - \frac{x}{{\log x}} + c$

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

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