MCQ
$\int_1^e {\frac{{1 + \log x}}{x}\,dx = } $
  • $\frac{3}{2}$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{e}$
  • D
    None of these

Answer

Correct option: A.
$\frac{3}{2}$
a
(a) $I = \int_1^e {\frac{{1 + \log x}}{x}\,} dx = \int_1^e {\frac{1}{x}dx + \int_1^e {\frac{{\log x}}{x}dx} } $

==> $[{\log _e}x]_1^e + \left[ {\frac{{{{(\log x)}^2}}}{2}} \right]_1^e = \frac{3}{2}$.

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