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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