A) 2
B) 1
C) \[-1\]
D) 0
Correct Answer: D
Solution :
Let \[f(x)=\log \left( \frac{2-x}{2+x} \right)\] \[\Rightarrow f(-x)=\log {{\left( \frac{2-x}{2+x} \right)}^{-1}}=-\log \left( \frac{2-x}{2+x} \right)=-f(x)\] \[\therefore \] \[\int_{-1}^{1}{\log \left( \frac{2-x}{2+x} \right)\,\,dx=0}\].You need to login to perform this action.
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