TITLE: Help proving an inequality in calculus $0 \leq \frac{x\ln(x)}{x^2-1}\leq \frac{1}{2}$
QUESTION [5 upvotes]: So as the title states, I need help with proving this inequality:
$$0 \leq \frac{x\ln(x)}{x^2-1}\leq \frac{1}{2} $$
Also an important thisn is that $x>1$.
So I thought I could look different kinds fo x, but it's a subject of calculus, so perhaps it's related to Lagrange's or Rolle's theorem, but i cannot get an idea, i tried changing the inequality but it's not really working, so any help with solution would be really appreciated.
Thank you in advance.

REPLY [2 votes]: Bounding the integral by a trapezoid on the interval $[1,x]$ we get for $x>1$
$$\int_1^x\frac{\mathrm{d}t}t < \tfrac12(1+\frac1x)(x-1) = \frac{x^2-1}{2x}.$$