<s>[INST] Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le \int_5^8 \left(1-\frac{12}{2x+5}\right)dx\le\frac{9}{7}$$ Whilee the first one could be reduced to what is needed, how can I deal with the right bound?
 [/INST] $f(x)=\frac{2x-7}{2x+5}$ is a concave function on $[5,8]$, hence by the Hermite-Hadamard inequality the wanted integral is bounded between 
$$ \frac{1}{2}f(5)+f(6)+f(7)+\frac{1}{2}f(8) = \frac{11043}{11305} > \frac{42}{43}$$
and 
$$ f\left(\tfrac{11}{2}\right)+f\left(\tfrac{13}{2}\right)+f\left(\tfrac{15}{2}\right) = \frac{59}{60}<1.$$
As an alternative, one may notice that the exact value is given by $3-6\log\left(1+\frac{2}{5}\right)$, and by Padé approximants $\log(x+1)$ is bounded between $\frac{x}{1+\frac{x}{2}}$ and $\frac{x+\frac{x^2}{6}}{1+\frac{2x}{3}}$ over $[0,1]$. This gives that the wanted integral is between $\frac{93}{95}$ and $1$.
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