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