Integral Calculus Question 336
Question: $ \int_{{}}^{{}}{[ \frac{1}{\log x}-\frac{1}{{{(\log x)}^{2}}} ]dx=} $
Options:
A) $ \frac{1}{\log x}+c $
B) $ \frac{x}{\log x}+c $
C) $ \frac{x}{{{(\log x)}^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{[ \frac{1}{\log x}-\frac{1}{{{(\log x)}^{2}}} ]},dx=\int_{{}}^{{}}{\frac{1}{\log x},dx-\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}},dx}} $ $ =\frac{x}{\log x}+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}},.,\frac{1}{x}x,dx}-\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}}dx+c=\frac{x}{\log x}+c $ .
BETA
