Answer
\[ = \ln \,\,\left| {\frac{{\sqrt {1 + 4{e^t}} - 1}}{{\sqrt {1 + 4{e^t}} + 1}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dt}}{{\sqrt {1 + 4{e^t}} }}} \hfill \\
\hfill \\
Rewrite\,\,\, \hfill \\
\hfill \\
= \frac{1}{4}\int_{}^{} {\frac{{4{e^t}dt}}{{{e^t}\sqrt {1 + 4{e^t}} }}} \hfill \\
\hfill \\
set\,\,u = 4{e^t}\,\,\,\,then\,\,\,\,du = 4{e^t}dt \hfill \\
\hfill \\
= \int_{}^{} {\frac{{du}}{{u\sqrt {1 + u} }}} \hfill \\
\hfill \\
integrate\,\,\,by\,\,tables \hfill \\
\hfill \\
= \frac{1}{{\sqrt 1 }}\ln \,\,\left| {\frac{{\sqrt {1 + u} - 1}}{{\sqrt {1 + u} + 1}}} \right| + C \hfill \\
\hfill \\
substitute\,\,back\,\,u = 4{e^t}\,\,and\,\,simplify \hfill \\
\hfill \\
= \ln \,\,\left| {\frac{{\sqrt {1 + 4{e^t}} - 1}}{{\sqrt {1 + 4{e^t}} + 1}}} \right| + C \hfill \\
\end{gathered} \]
