Integral Calculus Question 381
Question: $ \int_{{}}^{{}}{\frac{1}{x{{(\log x)}^{2}}}}\ dx= $
Options:
A) $ \frac{1}{\log x}+c $
B) $ -\frac{1}{\log x}+c $
C) $ \log \log x+c $
D) $ -\log \log x+c $
Show Answer
Answer:
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. $
BETA
