On $(0,1)$ its absolute value is at most $x|\log x|$, which has integral $1/4$, by <integration by parts> or the substitution $x=e^{-t}$. The function is measurable and this majorant is <integrable>, so $\boxed{x\log x/(1+x)^2\text{ is integrable on }(0,1)}$.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D / D3 / ii / a / Solution

 Solution (source code)