Using a Table of Integrals with the appropriate substitution, find
e3x+4e2x+5exe2x16 dx\displaystyle \int\frac{{{e}^{{{3}{x}}}+{4}{e}^{{{2}{x}}}+{5}{e}^{{x}}}}{{\sqrt{{{e}^{{{2}{x}}}-{16}}}}}\ {\left.{d}{x}\right.}


Answer:
+C\displaystyle +{C}