đặt $\begin{cases}\ln x+1=u \\ x^{2}dx=dv \end{cases} \Rightarrow \begin{cases}du=\frac{1}{x+1}dx \\ v= \frac{1}{3}(x^{3}+1)\end{cases}$$\int\limits x^{2}\ln x+1=\frac{1}{3}(x^{3}+1)\ln (x+1)-\frac{1}{3} \int\limits (x^{2}-x+1)=\frac{1}{3}(x^{3}+1)\ln (x+1)-\frac{1}{3}(\frac{1}{3}x^{3}-\frac{1}{2}x^{2}+x)$