If $$y = \frac{{2 - x}}{{1 + x}},$$   then what is the value of $$\frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}}?$$

$$\eqalign{ & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{{y - 1 + 2y + 1}}{{{y^2} - 1}} \cr & = \frac{{3y}}{{{y^2} - 1}} \cr & = \frac{{3\left( {\frac{{2 - x}}{{1 + x}}} \right)}}{{{{\left( {\frac{{2 - x}}{{1 + x}}} \right)}^2} - 1}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{4 + {x^2} - 4x - 1 - {x^2} - 2x}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{ - 6x + 3}} \cr & = \frac{{3\left( {2 - x} \right) \times \left( {1 + x} \right)}}{{3\left( {1 - 2x} \right)}} \cr & = \frac{{\left( {2 - x} \right)\left( {1 + x} \right)}}{{\left( {1 - 2x} \right)}} \cr & \cr & {\bf{Alternate}}\,{\bf{solution:}} \cr & y = \frac{{2 - x}}{{1 + x}} \cr & {\text{Put }}x = 0, \cr & y = 2 \cr & \frac{1}{{y + 1}} + \frac{{2y + 1}}{{{y^2} - 1}} \cr & = \frac{1}{3} + \frac{5}{3} \cr & = \frac{6}{3} \cr & = 2 \cr & {\text{Go through option}} \cr & \frac{{\left( {1 + x} \right)\left( {2 - x} \right)}}{{\left( {1 - 2x} \right)}} \cr & = \frac{{1 \times 2}}{1} \cr & = 2 \cr} $$