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

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 $\boxed{x\log x/(1+x{)}^2\text{ is integrable on }(0,1)}$.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, so $x\log x/(1+x{)}^2\ge (\log x)/(4x)\ge 0$. The latter integral diverges because its primitive is $(\log x{)}^2/8$. Thus the absolute integral diverges and the answer is $\boxed{\text{not integrable on }(1,\infty )}$.