If f(x)=ln⁡1+x1−x, then then f2x1+x2=

If $f (x) = \ln (\frac{1 + x}{1 - x}),$ then $then f (\frac{2 x}{1 + x^{2}}) =$

A
$f (x)$
B
$f (\frac{1}{x})$
C
$2 f (x)$
D
$2 f (\frac{1}{x})$

Solution:

$f (x) = \ln (\frac{1 + x}{1 - x}) \Rightarrow f (\frac{2 x}{1 + x^{2}}) = \ln (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$

$= \ln (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x}) = \ln {(\frac{1 + x}{1 - x})}^{2} = 2 \ln (\frac{1 + x}{1 - x}) = 2 f (x)$

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