If I _1= _ e ^ e ^2 d x x and I _2= _1^2 e ^x x d x, then

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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