$\begin{align}  & \int_{1}^{2}{x\log ({{x}^{2}}+1)dx} \\ & \text{Let }u={{x}^{2}}+1 \\ & \frac{du}{dx}=2x,\text{ }dx=\frac{du}{2x} \\ & \int{x\log ({{x}^{2}}+1)dx}=\int{x(\log (u)(\tfrac{du}{2x})}=\frac{1}{2}\int{\log udu} \\ & \text{Using integration by part for }\frac{1}{2}\int{\log udu} \\ & \frac{1}{2}\int{\log udu}=\frac{1}{2}\left[ u\log u-\int{u\cdot \frac{1}{u}}du \right]=\frac{1}{2}\left[ u\log u-\int{du} \right] \\ & \frac{1}{2}\int{\log udu}=\frac{1}{2}\left[ u\log u-u \right] \\ & \int_{0}^{2}{x\log ({{x}^{2}}+1)dx}=\frac{1}{2}\left[ ({{x}^{2}}+1)\log ({{x}^{2}}+1)-({{x}^{2}}+1) \right]_{0}^{2} \\ & \int_{0}^{2}{x\log ({{x}^{2}}+1)dx}=\frac{1}{2}\left[ (5\log 5-5)-(2\log 2-2) \right] \\ & \int_{0}^{2}{x\log ({{x}^{2}}+1)dx}=\frac{1}{2}[5\log 5-5-2\log 2+2] \\ & \int_{0}^{2}{x\log ({{x}^{2}}+1)dx}=\frac{5}{2}\log 5-\log 2-\frac{3}{2} \\\end{align}$