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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{1}{{x{{(\log x)}^2}}}} \;dx = $
  • A
    $\frac{1}{{\log x}} + c$
  • ✓
    $ - \frac{1}{{\log x}} + c$
  • C
    $\log \log x + c$
  • D
    $ - \log \log x + c$
✓

Answer

Correct option: B.
$ - \frac{1}{{\log x}} + c$
b
(b) Put $\log x = t \Rightarrow \frac{1}{x}\,dx = dt,$ then
$\int_{}^{} {\frac{1}{{x{{(\log x)}^2}}}\,dx} = \int_{}^{} {\frac{1}{{{t^2}}}\,dt} = - \frac{1}{t} + c = - \frac{1}{{\log x}} + c.$

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