MCQ
$\int_{ - 1}^1 {\log \left( {\frac{{1 + x}}{{1 - x}}} \right)\,dx = } $
- A$2$
- B$1$
- ✓$0$
- D$\pi $
then $f( - x) = \log \left( {\frac{{1 - x}}{{1 + x}}} \right) = - f(x)$
Therefore, $\int_{ - 1}^1 {\log \left( {\frac{{1 + x}}{{1 - x}}} \right){\rm{ }}dx = 0} $.
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