If \(f\left( x \right) = lo{g_e}\left( {\frac{{1 - x}}{{1 + x}}} \right)\), |x| < 1, then \(f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\) is equal to:

If \(f\left( x \right) = lo{g_e}\left( {\frac{{1 - x}}{{1 + x}}} \righ

1.	If \(f\left( x \right) = lo{g_e}\left( {\frac{{1 - x}}{{1 + x}}} \right)\), \|x\| < 1, then \(f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\) is equal to:
A.	2f(x)
B.	2f(x2)
C.	(f(x))2
D.	-2f(x)
Answer» B. 2f(x2)

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