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.