Answer
$\displaystyle \frac{-1.1^{-x+4}+1.1^{x+4}}{\ln(1.1)}+C$
Work Step by Step
$\displaystyle \int(1.1^{-x+4}+1.1^{x+4})dx=\int(1.1^{-x+4})dx+\int(1.1^{x+4})dx$
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=1.1\\
a=-1\\
b=4
\end{array}\right]$ and $\left[\begin{array}{l}
c=1.1\\
a=1\\
b=4
\end{array}\right]$
$=\displaystyle \frac{1.1^{-x+4}}{(-1)\ln(1.1)}+\frac{1.1^{x+4}}{(1)\ln(1.1)}+C$
= $\displaystyle \frac{-1.1^{-x+4}+1.1^{x+4}}{\ln(1.1)}+C$
