Answer
\[ = 3\sqrt[3]{2}\]
Work Step by Step
\[\begin{gathered}
\int_0^{\ln 3} {\frac{{{e^y}}}{{\,{{\left( {{e^y} - 1} \right)}^{\frac{2}{3}}}}}\,\,dy} \hfill \\
\hfill \\
integration\,\,of\,\,indefinite\,\,integral \hfill \\
\hfill \\
u = {e^y}\,\,\,\,then\,\,\,du = {e^y}dy\, \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^y}}}{{\,{{\left( {{e^y} - 1} \right)}^{\frac{2}{3}}}}}} \,dy = \int_{}^{} {\frac{1}{{\,{{\left( {u - 1} \right)}^{\frac{2}{3}}}}}\,du} \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
= 3\,{\left( {u - 1} \right)^{\frac{1}{3}}} + C \hfill \\
\hfill \\
= 3\,{\left( {{e^y} - 1} \right)^{\frac{1}{3}}} + C \hfill \\
\hfill \\
use\,\,\int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\
\hfill \\
= 2\,\left( {e - 1} \right)\int_0^{\ln 3} {\frac{{{e^y}}}{{\,{{\left( {{e^y} - 1} \right)}^{\frac{2}{3}}}}}\,\,dy} \,\, = \,\,\mathop {\lim }\limits_{c \to {0^ + }} \int_c^{\ln 3} {\frac{{{e^y}}}{{\,{{\left( {{e^y} - 1} \right)}^{\frac{2}{3}}}}}\,\,dy} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \,\mathop {\lim }\limits_{c \to {0^ + }} \,\,\,\,\left[ {3\,{{\left( {{e^y} - 1} \right)}^{\frac{1}{3}}}} \right]_c^{\ln 3} \hfill \\
\hfill \\
use\,\,\,the\,\,ftc \hfill \\
\hfill \\
= 3\,\,\left( {\,{{\left( {3 - 1} \right)}^{\frac{1}{3}}} - \mathop {\lim }\limits_{c \to {0^ + }} \,{{\left( {{e^c} - 1} \right)}^{\frac{1}{3}}}} \right) \hfill \\
\hfill \\
{\text{evaluate}}\,\,{\text{the}}\,\,{\text{limit}} \hfill \\
\hfill \\
= 3\,\left( {\,{{\left( {3 - 1} \right)}^{\frac{1}{3}}} - \,{{\left( {{e^0} - 1} \right)}^{\frac{1}{3}}}} \right) \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= 3\sqrt[3]{2} \hfill \\
\end{gathered} \]
