Tính tích phân \(I=\int_{0}^{1}(2x+1)e^xdx\) Đặt \(\left\{\begin{matrix} u=2x+1\\ dv=e^xdx \end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2dx\\ v=e^x \end{matrix}\right.\) Do đó \(I=\int_{0}^{1}(2x+1)e^xdx=(2x+1)e^x\bigg|^1_0-\int_{0}^{1}2x^xdx\) \(=[(2x+1)e^x-2x^x]\bigg|^1_0-\int_{0}^{1}=e+1\)