## Thomas' Calculus 13th Edition

$$\frac{1}{{5\left( {z - 3} \right)}} - \frac{1}{{5\left( {z + 2} \right)}}$$
\eqalign{ & \frac{z}{{{z^3} - {z^2} - 6z}} \cr & {\text{Factoring the denominator}} \cr & = \frac{z}{{z\left( {{z^2} - z - 6} \right)}} \cr & = \frac{z}{{z\left( {z - 3} \right)\left( {z + 2} \right)}} \cr & = \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} \cr & {\text{The partial fraction decomposition has the form}} \cr & \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{A}{{z - 3}} + \frac{B}{{z + 2}}\,\,\,\left( {\bf{1}} \right) \cr & {\text{Multiplying by }}z\left( {z - 3} \right)\left( {z + 2} \right){\text{, we have}} \cr & 1 = A\left( {z + 2} \right) + B\left( {z - 3} \right) \cr & \cr & \,\,\,\,\,{\text{set }}z = 3 \cr & 1 = A\left( {3 + 2} \right) + B\left( {3 - 3} \right) \cr & A = \frac{1}{5} \cr & \cr & \,\,\,\,\,{\text{set }}z = - 2 \cr & 1 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 3} \right) \cr & B = - \frac{1}{5} \cr & \cr & {\text{Then substituting }}A{\text{ and }}B{\text{ into the equation }}\left( {\bf{1}} \right) \cr & \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{{1/5}}{{z - 3}} + \frac{{ - 1/5}}{{z + 2}} \cr & \frac{1}{{\left( {z - 3} \right)\left( {z + 2} \right)}} = \frac{1}{{5\left( {z - 3} \right)}} - \frac{1}{{5\left( {z + 2} \right)}} \cr}