Integrate the function: $\int x(\log x)^{2} \, dx$

Let $I = \int x(\log x)^{2} \, dx$.
Using integration by parts,where $\int u \, dv = uv - \int v \, du$.
Let $u = (\log x)^{2}$ and $dv = x \, dx$.
Then $du = 2 \log x \cdot \frac{1}{x} \, dx$ and $v = \frac{x^{2}}{2}$.
$I = (\log x)^{2} \cdot \frac{x^{2}}{2} - \int \frac{x^{2}}{2} \cdot 2 \log x \cdot \frac{1}{x} \, dx$
$I = \frac{x^{2}}{2}(\log x)^{2} - \int x \log x \, dx$.
Now,integrate $\int x \log x \, dx$ by parts again.
Let $u = \log x$ and $dv = x \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^{2}}{2}$.
$\int x \log x \, dx = \log x \cdot \frac{x^{2}}{2} - \int \frac{x^{2}}{2} \cdot \frac{1}{x} \, dx$
$= \frac{x^{2}}{2} \log x - \frac{1}{2} \int x \, dx = \frac{x^{2}}{2} \log x - \frac{x^{2}}{4}$.
Substituting this back into the expression for $I$:
$I = \frac{x^{2}}{2}(\log x)^{2} - \left( \frac{x^{2}}{2} \log x - \frac{x^{2}}{4} \right) + C$
$I = \frac{x^{2}}{2}(\log x)^{2} - \frac{x^{2}}{2} \log x + \frac{x^{2}}{4} + C$.