## Calculus: Early Transcendentals (2nd Edition)

$= \ln \left| {\frac{{{z^3}\,\left( {z - 1} \right)}}{{\,{{\left( {z + 5} \right)}^3}}}} \right| + C$
$\begin{gathered} \int_{}^{} {\frac{{{z^2} + 20z - 15}}{{{z^3} + 4{z^2} - 5z}}} dz \hfill \\ \hfill \\ {\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\ \hfill \\ \frac{{{z^2} + 20z - 15}}{{{z^3} + 4{z^2} - 5z}} = \frac{{{z^2} + 20z - 15}}{{z\,\left( {{z^2} + 4z - 5} \right)\,}} \hfill \\ \hfill \\ Using\,\,partial\,fractions \hfill \\ \hfill \\ = \frac{{{z^2} + 20z - 15}}{{z\,\left( {z + 5} \right)\,\left( {z - 1} \right)}} = \frac{A}{z} + \frac{B}{{z + 5}} + \frac{C}{{z - 1}} \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ \hfill \\ A = 3,\,\,B = - 3,\,\,C = 1 \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{{{z^2} + 20z - 15}}{{z\,\left( {z + 5} \right)\,\left( {z - 1} \right)}} = - \frac{3}{{z + 5}} + \frac{3}{z} + \frac{1}{{z - 1}} \hfill \\ \hfill \\ \int_{}^{} {\frac{{{z^2} + 20z - 15}}{{{z^3} + 4{z^2} - 5z}}} dz = \int_{}^{} {\,\left( { - \frac{3}{{z + 5}} + \frac{3}{z} + \frac{1}{{z - 1}}} \right)} dz \hfill \\ \hfill \\ = \int_{}^{} {\,\left( { - \frac{3}{{z + 5}} + \frac{3}{z} + \frac{1}{{z - 1}}} \right)dz} \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ = - 3\ln \left| {z + 5} \right| + 3\ln \left| z \right| + \ln \left| {z - 1} \right| + C \hfill \\ \hfill \\ = - \ln \left| {\,{{\left( {z + 5} \right)}^3}} \right| + \ln \left| {{z^3}} \right| + \ln \left| {z - 1} \right| + C \hfill \\ \hfill \\ = \ln \left| {\frac{{{z^3}\,\left( {z - 1} \right)}}{{\,{{\left( {z + 5} \right)}^3}}}} \right| + C \hfill \\ \hfill \\ \end{gathered}$