Using a Table of Integrals with the appropriate substitution, find
e3x+5e2x+6exe2x100 dx\displaystyle \int\frac{{{e}^{{{3}{x}}}+{5}{e}^{{{2}{x}}}+{6}{e}^{{x}}}}{{\sqrt{{{e}^{{{2}{x}}}-{100}}}}}\ {\left.{d}{x}\right.}


Answer:
+C\displaystyle +{C}