Answer
$$6{e^x} + \frac{{{2^{x + 2}}}}{{\ln 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {6{e^x} + 4\left( {{2^x}} \right)} \right]} dx \cr
& {\text{sum rule for derivatives}} \cr
& = \int {6{e^x}} dx + \int {4\left( {{2^x}} \right)} dx \cr
& {\text{use the constant multiple rule }}\int {kf\left( x \right)dx} = k\int {f\left( x \right)} dx \cr
& = 6\int {{e^x}} dx + 4\int {{2^x}} dx \cr
& {\text{integrate}} \cr
& = 6\left( {{e^x}} \right) + 4\left( {\frac{{{2^x}}}{{\ln 2}}} \right) + C \cr
& = 6\left( {{e^x}} \right) + {2^2}\left( {\frac{{{2^x}}}{{\ln 2}}} \right) + C \cr
& {\text{simplifying}} \cr
& = 6{e^x} + \frac{{{2^{x + 2}}}}{{\ln 2}} + C \cr} $$
