A) \[\frac{3}{2}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{e}\]
D) None of these
Correct Answer: A
Solution :
\[I=\int_{1}^{e}{\frac{1+\log x}{x}\,}dx=\int_{1}^{e}{\frac{1}{x}dx+\int_{1}^{e}{\frac{\log x}{x}dx}}\] Þ \[[{{\log }_{e}}x]_{1}^{e}+\left[ \frac{{{(\log x)}^{2}}}{2} \right]_{1}^{e}=\frac{3}{2}\].
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