Using a Table of Integrals with the appropriate substitution, find
∫
e
3
x
+
4
e
2
x
+
5
e
x
e
2
x
−
16
d
x
\displaystyle \int\frac{{{e}^{{{3}{x}}}+{4}{e}^{{{2}{x}}}+{5}{e}^{{x}}}}{{\sqrt{{{e}^{{{2}{x}}}-{16}}}}}\ {\left.{d}{x}\right.}
∫
e
2
x
−
16
e
3
x
+
4
e
2
x
+
5
e
x
d
x
Answer:
+
C
\displaystyle +{C}
+
C
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