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For $x\in(0,1)$, let $$f(x):=\frac1{x\ln^2\frac ex};$$ this is the same function as then one suggested by Deane Yang. Then $f\ge0$ and $\int_0^1f(x)\,dx=1$. On the other hand, $\ln f(x)\sim\ln\frac ex$ as $x\downarrow0$. So, $\ln f(x)\ge\frac12\ln\frac ex$ for some $c\in(0,1)$ and all $x\in(0,c)$. So, $$\int_0^1 f(x)\ln f(x)\,dx\ge \int_0^c \frac1{x\ln^2\frac ex}\frac12\ln\frac ex\,dx=\infty,$$ as desired.