Integral Calculus Question 180
Question: $ \int_{{}}^{{}}{[ \log (\log x)+\frac{1}{{{(\log x)}^{2}}} ]}\ dx= $
Options:
A) $ x\log (\log x)+\frac{x}{\log x}+c $
B) $ x\log (\log x)-\frac{x}{\log x}+c $
C) $ x\log (\log x)+\frac{\log x}{x}+c $
D) $ x\log (\log x)-\frac{\log x}{x}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{[ \log (\log x)+\frac{1}{{{(\log x)}^{2}}} ],dx}=\int_{{}}^{{}}{\log (\log x)dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}}}dx $ $ =x\log (\log x)-\int_{{}}^{{}}{\frac{x}{x\log x},dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx}} $ $ =x\log (\log x)-\frac{x}{\log x}-\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx}} $ $ =x\log (\log x)-\frac{x}{\log x}+c. $
BETA
