\int_{1}^{e} x(lnx+1) \text{d}x= \int (lnx \times x+x)\text{d}x= \int (lnx\times x)\text{d}x+\int (x)\text{d}x= \frac {x^2}{2}\times lnx-\int (\frac {x^2}{2}\times\frac {1}{x})\text{d}x+\int (x)\text{d}x=\frac{x^2}{2}\times lnx-\frac {x^2}{4}+\frac{x^2}{2}