Answer
$\frac{x^{2}}{2}(\ln x)^{2}-\frac{x^{2} \ln x}{2}+\frac{x^{2}}{4}+C $
Work Step by Step
$$
\begin{aligned}
& \int x(\ln x)^{2} d x\\
& \text { } \\
& \text { "Integrate by parts" } \\
& u=(\ln x)^{2} \quad d v=x d x \\
& d u=2 \ln x \cdot \frac{1}{x} d x \quad v=\frac{x^{2}}{2} \\
& u \cdot v-\int v d u \\
& \text { } \\
& (\ln x)^{2} \frac{x^{2}}{2}-\int \frac{x^{2}}{2} \frac{2 \ln x}{x} d x \\
& (\ln x)^{2} \frac{x^{2}}{2}-\int x \ln x d x \\
& \text { } \\
& \text { "Integrate by parts again" } \\
& \quad u=\ln x \quad d v=x d x \\
& \quad d u=\frac{1}{x} d x \quad v=\frac{x^{2}}{2} \\
& \text { } \\
& (\ln x)^{2} \frac{x^{2}}{2}-\left[\ln x \cdot \frac{x^{2}}{2}-\int \frac{x^{2}}{2} \frac{1}{x} d x\right] \\
& (\ln x)^{2} \frac{x^{2}}{2}-\left[\frac{\ln x \cdot x^{2}}{2}-\int \frac{x}{2} d x\right] \\
& \frac{x^{2}}{2}(\ln x)^{2}-\left[\frac{x^{2} \ln x}{2}-\frac{x^{2}}{4}\right]+C \\
& \text { } \\
& \frac{x^{2}}{2}(\ln x)^{2}-\frac{x^{2} \ln x}{2}+\frac{x^{2}}{4}+C
\end{aligned}
$$
