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Question: \(\int _ { - 1 } ^ { 1 } \log \frac { 2 - x } { 2 + x } d x =\)...

∫−11log⁡2−x2+xdx=\int _ { - 1 } ^ { 1 } \log \frac { 2 - x } { 2 + x } d x =

A

2

B

1

C

−1- 1

D

0

Answer

0

Explanation

Solution

Let f(x)=log⁡(2−x2+x)f ( x ) = \log \left( \frac { 2 - x } { 2 + x } \right)

⇒f(−x)=log⁡(2−x2+x)−1=−log⁡(2−x2+x)=−f(x)\Rightarrow f ( - x ) = \log \left( \frac { 2 - x } { 2 + x } \right) ^ { - 1 } = - \log \left( \frac { 2 - x } { 2 + x } \right) = - f ( x )

∴\therefore ∫−11log⁡(2−x2+x)dx=0\int _ { - 1 } ^ { 1 } \log \left( \frac { 2 - x } { 2 + x } \right) d x = 0 .