## Calculus: Early Transcendentals (2nd Edition)

$= - 2\ln \,\left( {{e^t} + \sqrt {4 + {e^{2t}}} } \right) + \frac{{{e^t}\sqrt {4 + \,{e^{2t}}} }}{2} + C$
$\begin{gathered} \int_{}^{} {\frac{{{e^{3t}}}}{{\sqrt {4 + {e^{2t}}} }}} \hfill \\ \hfill \\ set \hfill \\ \hfill \\ u = {e^t}\,\,\,\,\,then\,\,\,\,\,du = {e^t}dt \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{{e^{3t}}}}{{\sqrt {4 + {e^{2t}}} }}} = \int_{}^{} {\frac{{{e^{2t}}{e^t}}}{{\sqrt 4 + {e^{2t}}}}dt} = \int_{}^{} {\frac{{\,{{\left( {{e^t}} \right)}^2}}}{{\sqrt {4 + \,{{\left( {{e^t}} \right)}^2}} }}dt} \hfill \\ \hfill \\ \int_{}^{} {\frac{{{e^{3t}}}}{{\sqrt {4 + {e^{2t}}} }}} \,dt\,\, = \int_{}^{} {\frac{{{u^2}}}{{\sqrt {4 + {u^2}} }}} \,du \hfill \\ \hfill \\ use\,\,\,the\,\,formula\,\int_{}^{} {\frac{{{x^2}}}{{\sqrt {{a^2} + {x^2}} }}\,dx} = - \frac{{{a^2}}}{2}\ln \,\left( {x + \sqrt {{a^2} + {x^2}} } \right) + \frac{{x\sqrt {{a^2} + {x^2}} }}{2} + C \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{{u^2}}}{{\sqrt {4 + {u^2}} }}} \,du = - \frac{{{4^2}}}{2}\ln \,\left( {u + \sqrt {4 + {u^2}} } \right) + \frac{{u\sqrt {4 + {u^2}} }}{2} + C \hfill \\ \hfill \\ substitute\,\,back\,\,u = {e^t}\, \hfill \\ \hfill \\ = - 2\ln \,\left( {{e^t} + \sqrt {4 + {e^{2t}}} } \right) + \frac{{{e^t}\sqrt {4 + \,{{\left( {{e^t}} \right)}^2}} }}{2} + C \hfill \\ \hfill \\ = - 2\ln \,\left( {{e^t} + \sqrt {4 + {e^{2t}}} } \right) + \frac{{{e^t}\sqrt {4 + \,{e^{2t}}} }}{2} + C \hfill \\ \hfill \\ \hfill \\ \hfill \\ \end{gathered}$