_1^2 1 x^2 e^ - 1 x ,dx = - Vidyadip

MCQ

$\int_1^2 {\frac{1}{{{x^2}}}{e^{\frac{{ - 1}}{x}}}\,dx = } $

A
$\sqrt e + 1$
B
$\sqrt e - 1$
C
$\frac{{\sqrt e + 1}}{e}$
✓
$\frac{{\sqrt e - 1}}{e}$

✓

Answer

Correct option: D.

$\frac{{\sqrt e - 1}}{e}$

d
(d) Put $t = - \frac{1}{x} \Rightarrow dt = \frac{1}{{{x^2}}}dx$,

then it reduces to

$\int_{ - 1}^{ - 1/2} {{e^t}dt = [{e^t}]_{ - 1}^{ - 1/2} = {e^{ - 1/2}} - {e^{ - 1}}} = \frac{{\sqrt e - 1}}{e}$.

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