MCQ
If $I _1=\int_{ e }^{ e ^2} \frac{d x}{\log x}$ and $I _2=\int_1^2 \frac{ e ^x}{x} d x$, then
  • $I _1= I _2$
  • B
    $I _1> I _2$
  • C
    $I _1< I _2$
  • D
    None of these

Answer

Correct option: A.
$I _1= I _2$
(A)
$I _1=\int_{ e }^{ e ^2} \frac{d x}{\log x}$
Put $\log x= t$
$\Rightarrow d x=x dt = e ^{ t } dt$
When $x= e , t =1$ and when $x= e ^2, t =2$
$\therefore \quad I _1=\int_1^2 \frac{ e ^{ t }}{ t } dt$
$=\int_1^2 \frac{ e ^x}{x} d x \quad \ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( t ) dt \right]$
$\therefore \quad I _1= I _2$

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