[MP PET 1995; Pb. CET 2000]
A) 2
B) 1
C) 0
D) $ \pi $
Correct Answer: C
If $ f(x)=\log ( \frac{1+x}{1-x} ), $ then $ f(-x)=\log ( \frac{1-x}{1+x} )=-f(x) $
Therefore, $ \int _{-1}^{1}{\log ( \frac{1+x}{1-x} )dx=0} $ .
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