A) \[{{e}^{-\log x}}+c\]
B) \[-x{{e}^{-\log x}}+c\]
C) \[{{e}^{\log x}}+c\]
D) \[\log |x|+c\]
Correct Answer: D
Solution :
Let\[I=\int{{{e}^{-\log x}}dx}\] \[=\int{{{e}^{\log \frac{1}{x}}}}dx\int{\frac{1}{x}dx}\] \[=\log |x|+c\]
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