A) \[\frac{1}{2}{{e}^{x}}-{{\log }_{e}}x+C\]
B) \[\frac{1}{2}{{\log }_{e}}x-{{e}^{x}}+C\]
C) \[\frac{1}{2}{{({{\log }_{e}}x)}^{2}}-x+C\]
D) \[\frac{{{e}^{x}}}{2x}+C\]
Correct Answer: C
Solution :
[c] \[\int{\frac{1}{x}}{{\log }_{e}}\left( \frac{x}{{{e}^{x}}} \right)dx\] \[=\int{\frac{1}{x}({{\log }_{e}}x-{{\log }_{e}}{{e}^{x}})\,dx}\] \[=\int{\frac{{{\log }_{e}}x-x}{x}dx}\] \[=\int{\frac{1}{x}{{\log }_{e}}x\,dx-\int{1\,dx}}\] \[=\frac{1}{2}{{({{\log }_{e}}x)}^{2}}-x+C\]
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