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

MCQ

$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = } $

A
$\frac{{{e^2}}}{2} + e$
B
$e - \frac{{{e^2}}}{2}$
✓
$\frac{{{e^2}}}{2} - e$
D
None of these

✓

Answer

Correct option: C.

$\frac{{{e^2}}}{2} - e$

c
(c)$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = \left[ {\frac{1}{x}{e^x}} \right]_1^2 = \frac{{{e^2}}}{2} - e} $.

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