已知\(f\left ( {\frac {1-x} {1+x}} \right )=\frac {1-{x}^{2}} {1+{x}^{2}}\)，则\(f\

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已知\(f\left ( {\frac {1-x} {1+x}} \right )=\frac {1-{x}^{2}} {1+{x}^{2}}\)，则\(f\left ( {x} \right )\)的解析式为(　　)

A.\(f\left ( {x} \right )=\frac {x} {1+{x}^{2}}\left ( {x≠-1} \right )\)

B.\(f\left ( {x} \right )=\frac {-2x} {1+{x}^{2}}\left ( {x≠-1} \right )\)

C.\(f\left ( {x} \right )=\frac {2x} {1+{x}^{2}}\left ( {x≠-1} \right )\)

D.\(f\left ( {x} \right )=\frac {-x} {1+{x}^{2}}\left ( {x≠-1} \right )\)

参考答案：C