A) \[\frac{1}{\log x}+c\]
B) \[-\frac{1}{\log x}+c\]
C) \[\log \log x+c\]
D) \[-\log \log x+c\]
Correct Answer: B
Solution :
Put \[\log x=t\Rightarrow \frac{1}{x}\,dx=dt,\] then \[\int_{{}}^{{}}{\frac{1}{x{{(\log x)}^{2}}}\,dx}=\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt}=-\frac{1}{t}+c=-\frac{1}{\log x}+c.\]
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