Integrate the function: $\left(x^{2}+1\right) \log x$

(N/A) Let $I = \int \left(x^{2}+1\right) \log x \, dx = \int x^{2} \log x \, dx + \int \log x \, dx$.
Let $I = I_{1} + I_{2}$ where $I_{1} = \int x^{2} \log x \, dx$ and $I_{2} = \int \log x \, dx$.
For $I_{1}$,using integration by parts with $\log x$ as the first function and $x^{2}$ as the second function:
$I_{1} = \log x \cdot \frac{x^{3}}{3} - \int \frac{1}{x} \cdot \frac{x^{3}}{3} \, dx = \frac{x^{3}}{3} \log x - \frac{1}{3} \int x^{2} \, dx = \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} + C_{1}$.
For $I_{2}$,using integration by parts with $\log x$ as the first function and $1$ as the second function:
$I_{2} = \log x \cdot x - \int \frac{1}{x} \cdot x \, dx = x \log x - \int 1 \, dx = x \log x - x + C_{2}$.
Combining these results:
$I = \left( \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} \right) + (x \log x - x) + C$
$I = \left( \frac{x^{3}}{3} + x \right) \log x - \frac{x^{3}}{9} - x + C$.