A) \[\frac{1}{4{{x}^{2}}}(2\log x-1)+c\]
B) \[-\frac{1}{4{{x}^{2}}}(2\log x+1)+c\]
C) \[\frac{1}{4{{x}^{2}}}(2\log x+1)+c\]
D) \[\frac{1}{4{{x}^{2}}}(1-2\log x)+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{\log x}{{{x}^{3}}}dx=\int_{{}}^{{}}{{{x}^{-3}}\log x\ dx}}\] \[=-\frac{\log x}{2{{x}^{2}}}+\int_{{}}^{{}}{\frac{1}{x}.\frac{1}{2{{x}^{2}}}+c=-\frac{\log x}{2{{x}^{2}}}+\frac{1}{2}.\frac{{{x}^{-2}}}{-2}+c}\] \[=-\frac{\log x}{2{{x}^{2}}}-\frac{1}{4{{x}^{2}}}+c=-\frac{1}{4{{x}^{2}}}(2\log x+1)+c\].
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