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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
$\int_{}^{} {\frac{{\log x}}{{{{(1 + \log x)}^2}}}dx = } $
  • A
    $\frac{1}{{1 + \log x}} + c$
  • B
    $\frac{x}{{{{(1 + \log x)}^2}}} + c$
  • ✓
    $\frac{x}{{1 + \log x}} + c$
  • D
    $\frac{1}{{{{(1 + \log x)}^2}}} + c$
✓

Answer

Correct option: C.
$\frac{x}{{1 + \log x}} + c$
c
(c)$\int_{}^{} {\frac{{\log x}}{{{{(1 + \log x)}^2}}}\,dx} $.

Put $1 + \log x = t \Rightarrow \frac{1}{x}dx = dt$

$ \Rightarrow dx = x\,dt = {e^{t - 1}}dt,$ then it reduces to

$\int_{}^{} {\frac{{(t - 1)\,{e^{t - 1}}}}{{{t^2}}}dt} = \int_{}^{} {{e^{t - 1}}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)\,dt} = \frac{{{e^{t - 1}}}}{t} = \frac{x}{{1 + \log x}} + c$.

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