MCQ
$\int_{}^{} {\frac{{x - 2}}{{x(2\log x - x)}}dx} = $
- A$\log (2\log x - x) + c$
- ✓$\log \left( {\frac{1}{{2\log x - x}}} \right) + c$
- C$\log (x - 2\log x) + c$
- D$\log \left( {\frac{1}{{x - 2\log x}}} \right) + c$
✓
Answer
Correct option: B.
$\log \left( {\frac{1}{{2\log x - x}}} \right) + c$
b
(b)$\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 $(2\log x - x) = t \Rightarrow \left( {\frac{2}{x} - 1} \right)\,dx = dt,$
(b)$\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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