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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