Answer
$$\frac{1}{3}\ln \left| {{e^{3z}} - 4} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{2z}}}}{{{e^{2z}} - 4{e^{ - z}}}}dz} \cr
& {\text{multiply }}\frac{{{e^{2z}}}}{{{e^{2z}} - 4{e^{ - z}}}}{\text{ by }}\frac{{{e^z}}}{{{e^z}}} \cr
& = \int {\frac{{{e^z}{e^{2z}}}}{{{e^z}{e^{2z}} - 4{e^z}{e^{ - z}}}}dz} \cr
& = \int {\frac{{{e^{3z}}}}{{{e^{3z}} - 4}}dz} \cr
& {\text{substitute }}u = {e^{3z}} - 4,{\text{ }}du = 3{e^{3z}}dz \cr
& = \int {\frac{{{e^{3z}}}}{{{e^{3z}} - 4}}dz} = \frac{1}{3}\int {\frac{{du}}{u}} \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{3}\ln \left| u \right| + C \cr
& {\text{ with}}\,\,\,u = {e^{3z}} - 4 \cr
& = \frac{1}{3}\ln \left| {{e^{3z}} - 4} \right| + C \cr} $$
