Answer
\[\ln \left| {{e^x} + \sqrt {{e^{2x}} + 4} } \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{e^x}}}{{\sqrt {{e^{2x}} + 4} }}} \hfill \\
\hfill \\
rewrite\,\,the\,\,{\text{denominator}} \hfill \\
\hfill \\
= \int_{}^{} {\frac{{{e^x}dx}}{{\sqrt {\,{{\left( {{e^x}} \right)}^2} + {2^2}} }}} \hfill \\
\hfill \\
Use\,\,the\,\,formula\,\,\,\int_{}^{} {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }} = } \ln \left| {u + \sqrt {{u^2} + {a^2}} } \right| + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{{e^x}}}{{\sqrt {{e^{2x}} + 4} }}} = \ln \left| {{e^x} + \sqrt {{e^{2x}} + 4} } \right| + C \hfill \\
\end{gathered} \]