A) \[{{e}^{e}}\]
B) \[{{e}^{e}}-e\]
C) \[{{e}^{e}}+e\]
D) None of these
Correct Answer: A
Solution :
\[\int_{1}^{e}{\frac{{{e}^{x}}}{x}(1+x\log x)dx=\int_{1}^{e}{\frac{1}{x}{{e}^{x}}dx}}\]\[+\int_{1}^{e}{{{e}^{x}}{{\log }_{e}}x\,\,dx}\] = \[[{{e}^{x}}\log x]_{1}^{e}-\int_{1}^{e}{{{e}^{x}}\log x\,dx+\int_{1}^{e}{{{e}^{x}}\log x\,dx}}\] = \[[{{e}^{e}}\log e-{{e}^{1}}{{\log }_{e}}1]={{e}^{e}}\].
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