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Let $I=\int x^{2} \log x d x$

Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtain

$\begin{aligned} I &=\log x \int x^{2} d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x^{2} d x\right\} d x \\ &=\log x\left(\frac{x^{3}}{3}\right)-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ &=\frac{x^{3} \log x}{3}-\int \frac{x^{2}}{3} d x \\ &=\frac{x^{3} \log x}{3}-\frac{x^{3}}{9}+\mathrm{C} \end{aligned}$