<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] We wish to show that $$\int_5^8{\frac{2x-7}{2x+5}dx}<1.$$ Let $g:(0,\infty)\to\mathbf R$ be given by $$g(x)=1-\int_5^x{1-\frac{12}{2w+5}dw}.$$ Then $$g'(x)={12\over{2x+5}}-1$$ and $$g''(x)={-24\over{(2x+5)^2}}<0\,\, \forall x.$$ Thus, $g$ is concave. Also, $g'=0$ at $x=3.5$, so that $g$ has its turning point $\left( g'(3)>0\,\text{and}\,g'(4)<0 \right)$ at $x=3.5$, where $g=2.5-6\log \frac 54>0$ since $4e^{5/12}>5.$ Thus, $g$ has just two roots.
Clearly, one of the roots must be $<3.5$, so that we need only consider the other root $>3.5$ (since we're interested only in how this function behaves in $[5,8]$). I claim that the other root is also $>8.$ Now we have that $$g(x)=6-x-6\log(2x+5),$$ so that $$g(8)=6\log \frac75 -2>0.$$ That last inequality holds because clearly $7>5e^{1/3}.$ Also, $$g(20)=-14+6\log 3<-14+12=-2<0,$$ since $3<e^2=(2+\varepsilon)^2=4+\delta$, where $1/2<\varepsilon<1$ and $\delta=4\varepsilon+\varepsilon^2.$
Consequently, the other root of $g$ must be in $(8,20)$, so that $g>0$ in $[5,8]$ (since $g(5)=1>0$). The result follows by setting $x=8$ in $g(x)>0.\blacksquare$
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