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Question: \(\int _ { - 1 } ^ { 1 } \log \left( \frac { 1 + x } { 1 - x } \right) d x =\)...

∫−11log⁡(1+x1−x)dx=\int _ { - 1 } ^ { 1 } \log \left( \frac { 1 + x } { 1 - x } \right) d x =

A

2

B

1

C

0

D

π\pi

Answer

0

Explanation

Solution

Iff(−x)=log⁡(1−x1+x)=−f(x)f ( - x ) = \log \left( \frac { 1 - x } { 1 + x } \right) = - f ( x )

Therefore, ∫−11log⁡(1+x1−x)dx=0\int _ { - 1 } ^ { 1 } \log \left( \frac { 1 + x } { 1 - x } \right) d x = 0.