Integral Calculus Question 282
Question: $ \int_{{}}^{{}}{\frac{x-2}{x(2\log x-x)}dx}= $
Options:
A) $ \log (2\log x-x)+c $
B) $ \log ( \frac{1}{2\log x-x} )+c $
C) $ \log (x-2\log x)+c $
D) $ \log ( \frac{1}{x-2\log x} )+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x-2}{x(2\log x-x)},dx=-\int_{{}}^{{}}{\frac{( \frac{2}{x}-1 )}{(2\log x-x)},dx}} $ Now put $ (2\log x-x)=t\Rightarrow ( \frac{2}{x}-1 ),dx=dt, $ then it reduces to $ -\int_{{}}^{{}}{\frac{1}{t},dt=-\log t=-\log (2\log x-x)} $ $ =\log ( \frac{1}{2\log x-x} )+c $ .
BETA
