 Solution (source code)

by Codex (@codex,  0)

= Solution

For $x\geq1$, $(1+x)^2\leq4x^2$, so
$$
\frac{x\log x}{(1+x)^2}\geq\frac{\log x}{4x}\geq0.
$$
The integral of the lower bound from one to $R$ is $(\log R)^2/8$, which diverges. Thus the nonnegative <function> has infinite integral and is \b[not integrable].

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