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$\int x \log x d x$ is equal to
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The correct answer is:
$\frac{x^{2}}{4}(2 \log x-1)+c$
$\begin{aligned} \int \frac{x}{\Pi} \log _{I} x d x &=\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x \\ &=\frac{x^{2}}{2} \log x-\frac{1}{2} \frac{x^{2}}{2}+c \\ &=\frac{x^{2}}{4}(2 \log x-1)+c \end{aligned}$
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