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$f(x) + f\left( {\frac{1}{{1 - x}}} \right) = x,\forall x \in \mathbb{R}$

$f(x) = ?$

$x \to 1 - \frac{1}{x}$ yields,

$f\left( {1 - \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 - \left( {1 - \frac{1}{x}} \right)}}} \right] = 1 - \frac{1}{x}$

$ \Rightarrow f\left( {\frac{{x - 1}}{x}} \right) + f(x) = \frac{{x - 1}}{x}........(A)$

$x \to \frac{1}{{1 - x}}$ yields,

$f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{1}{{1 - \frac{1}{{1 - x}}}}} \right) = \frac{1}{{1 - x}}$

$ \Rightarrow f\left( {\frac{1}{{1 - x}}} \right) + f\left( {\frac{{x - 1}}{x}} \right) = \frac{1}{{1 - x}}........(B)$

(A) - (B) yields,

$f(x) - f\left( {\frac{1}{{1 - x}}} \right) = \frac{{x - 1}}{x} - \frac{1}{{1 - x}}........(C)$

(C) + The original equation yields,

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

$ \Rightarrow f(x) = \frac{{{x^2} - {x^3} - 1 - {x^2} + 2x - x}}{{2x(1 - x)}} = \frac{{ - {x^3} + x - 1}}{{2x(1 - x)}}$