MCQ
$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = } $
- A$\frac{{{e^2}}}{2} + e$
- B$e - \frac{{{e^2}}}{2}$
- ✓$\frac{{{e^2}}}{2} - e$
- DNone 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} $.
(c)$\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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