## Calculus (3rd Edition)

$$\frac{2^xe^{4x}}{4+\ln 2}+c.$$
Since $2^x=e^{x\ln 2}$, then we have $$\int 2^xe^{4x}dx =\int e^{(4+\ln 2)x}dx=\frac{e^{(4+\ln 2)x}}{4+\ln 2}+c\\ =\frac{2^xe^{4x}}{4+\ln 2}+c.$$