Answer
$ \displaystyle \frac{1.1^{x}}{\ln 1.1}+\frac{2^{x}}{\ln 2}+C$
Work Step by Step
applying Sum and Difference RuIes,
... $=\displaystyle \int 1.1^{x}dx+\int 2^{x}dx$
... both integrals: $\displaystyle \int b^{x}dx=\frac{b^{x}}{\ln b}+C$
$=\displaystyle \frac{1.1^{x}}{\ln 1.1}+\frac{2^{x}}{\ln 2}+C$
