MCQ
$\int_1^e \frac{1+\log x}{x} d x=$
  • $\frac{3}{2}$
  • B
    $\frac{1}{7}$
  • C
    $\frac{1}{e}$
  • D
    $\frac{2}{3}$

Answer

Correct option: A.
$\frac{3}{2}$
(A)
Put $1+\log x= t \Rightarrow \frac{1}{x} d x= dt$
When $x=1, t =1$ and when $x= e , t =2$
$\therefore \quad \int_1^{ e } \frac{1+\log x}{x} d x=\int_1^2 tdt =\left[\frac{ t ^2}{2}\right]_1^2=\frac{3}{2}$

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