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Let $I=\int x \log 2 x d x$

Taking log 2x as first function and x as second function and integrating by parts, we obtain

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

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

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

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