答案:2{}^{x}e{}^{x}\dfrac{1}{1+\ln 2}解析:由分部积分法可得2^{x}e^{x}dx=\int 2^{x}de^{x}=2^{x}={2}^{x}{e}^{x}-\int {e}^{x}{2}^{x}\ln 2dx={2}^{x}{e}^{x}-\ln 2\int {e}^{x}{2}^{x}dx,\therefore \(1+\ln 2)\\int e^{x}2^{x}dx=2^{x}e^{x},\therefore \int {2}^{x}{e}^{x}dx={2}^{x}{e}^{x}\dfrac{1}{1+\ln 2}故