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