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

MCQ

$\int_1^e {\frac{{{e^x}}}{x}(1 + x\log x)\,dx} = $

✓
${e^e}$
B
${e^e} - e$
C
${e^e} + e$
D
None of these

✓

Answer

Correct option: A.

${e^e}$

a
(a) $\int_1^e {\frac{{{e^x}}}{x}(1 + x\log x)dx = \int_1^e {\frac{1}{x}{e^x}dx} } $$ + \int_1^e {{e^x}{{\log }_e}x\,\,dx} $

$=[{e^x}\log x]_1^e - \int_1^e {{e^x}\log x\,dx + \int_1^e {{e^x}\log x\,dx} } $

$= [{e^e}\log e - {e^1}{\log _e}1] = {e^e}$.

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