Answer
$ \displaystyle \frac{2^{3x+4}-2^{-3x+4}}{3\ln 2}+C$
Work Step by Step
$\displaystyle \int(2^{3x+4}+2^{-3x+4})dx=\int(2^{3x+4})dx+\int(2^{-3x+4})dx$
Shortcut formula:
$\displaystyle \qquad\int c^{ax+b}dx=\frac{\mathrm{l}}{a\ln c}\cdot c^{ax+b}+C$
Apply with $ \left[\begin{array}{l}
c=2\\
a=3\\
b=4
\end{array}\right]$ and $\left[\begin{array}{l}
c=2\\
a=-3\\
b=4
\end{array}\right]$
$= \displaystyle \frac{2^{3x+4}}{3\ln 2}+\frac{2^{-3x+4}}{-3\ln 2}+C$
=$ \displaystyle \frac{2^{3x+4}-2^{-3x+4}}{3\ln 2}+C$
