## Calculus: Early Transcendentals (2nd Edition)

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