Ta có \(I=\int_{0}^{1}xe^xdx+\int_{0}^{1}\frac{2x}{x+1}dx\) \(\int_{0}^{1}xe^x=\int_{0}^{1}xde^x=xe^x\bigg |^1_0-\int_{0}^{1}e^xdx=e-e^x=\bigg |^1_0=1\) \(\int_{0}^{1}\frac{2x}{x+1}dx=\int_{0}^{1}\left ( 2-\frac{2}{x+1} \right )dx=(2x-2ln\left | x+1 \right |) \bigg |^1_0=2-2ln2\) Do đó \(I=3-2ln2\)