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Solution:

Let $I=\int\left(x^{2}+1\right) \log x d x=\int x^{2} \log x d x+\int \log x d x$

Let $I=I_{1}+I_{2} \ldots$  ...(1)

Where, $I_{1}=\int x^{2} \log x d x$ and $I_{2}=\int \log x d x$

$I_{1}=\int x^{2} \log x d x$

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

$I_{1}=\log x-\int x^{2} d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x^{2} d x\right\} d x$

$=\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x$

$=\frac{x^{3}}{3} \log x-\frac{1}{3}\left(\int x^{2} d x\right)$

$=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+\mathrm{C}_{1}$   ...(2)

$I_{2}=\int \log x d x$

Taking log x as first function and 1 as second function and integrating by parts, we obtain

$I_{2}=\log x \int 1 \cdot d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int 1 \cdot d x\right\}$

$=\log x \cdot x-\int \frac{1}{x} \cdot x d x$

$=x \log x-\int 1 d x$

$=x \log x-x+\mathrm{C}_{2}$    ...(3)

Using equations (2) and (3) in (1), we obtain

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