A) \[2\log \frac{1-x}{1+x}\]
B) \[2\log \frac{1+x}{1-x}\]
C) \[\frac{1}{2}\log \frac{1+x}{1-x}\]
D) \[\frac{1}{2}\log \frac{1-x}{1+x}\]
Correct Answer: C
Solution :
\[{{\tanh }^{-1}}x=\frac{1}{2}\log \frac{1+x}{1-x}\]
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