Answer
$$ - \frac{2}{z} - \frac{1}{{{z^2}}} + \frac{2}{{z - 1}}$$
Work Step by Step
$$\eqalign{
& \frac{{z + 1}}{{{z^2}\left( {z - 1} \right)}} \cr
& {\text{The partial fraction decomposition has the form}} \cr
& \frac{{z + 1}}{{{z^2}\left( {z - 1} \right)}} = \frac{A}{z} + \frac{B}{{{z^2}}} + \frac{C}{{z - 1}}\,\,\,\left( {\bf{1}} \right) \cr
& {\text{Multiplying by }}{z^2}\left( {z - 1} \right){\text{, we have}} \cr
& z + 1 = Az\left( {z - 1} \right) + B\left( {z - 1} \right) + C{z^2} \cr
& z + 1 = A{z^2} - Az + Bz - B + C{z^2} \cr
& z + 1 = \left( {A{z^2} + C{z^2}} \right) + \left( { - Az + Bz} \right) - B \cr
& {\text{If we equate coefficients}}{\text{, we get the system}} \cr
& A + C = 0,\,\,\,\,\,\,\,\,\,\,\,\, - A + B = 1,\,\,\,\,\,\,\,\,\,\, - B = 1,\,\,\,\,\,\,\,B = - 1 \cr
& A = B - 1,\,\,\,A = - 2 \cr
& C = - A,\,\,\,\,C = 2 \cr
& {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr
& \frac{{z + 1}}{{{z^2}\left( {z - 1} \right)}} = - \frac{2}{z} - \frac{1}{{{z^2}}} + \frac{2}{{z - 1}} \cr} $$
