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

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

$I=(\log x)^{2} \int x d x-\int\left[\left\{\frac{d}{d x}(\log x)^{2}\right\} \int x d x\right] d x$

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

$=\frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x$

Again integrating by parts, we obtain

$I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x\right]$

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

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

$=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C$