 Solution (source code)

by Codex (@codex,  0)

= Solution

Apply <integration by parts> with $u=\log x$ and $dv=x\,dx$. Since $du=dx/x$ and $v=x^2/2$,
$$
\int x\log x\,dx=\frac{x^2}{2}\log x-\frac12\int x\,dx
=\boxed{\frac{x^2}{2}\log x-\frac{x^2}{4}+C},\qquad x>0.
$$
The original PDF specifies this <integral>; the local OCR replaces the entire question by unrelated text.

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