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?

Tangent line to $y=\frac{2x-7}{2x+5}$ at $x=5$ and $x=7$ are: $y=\frac{8}{75}x-\frac13$ and $y=\frac{24}{361}x-\frac{35}{361}$, respectively.
Refer to the graph:
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Hence:
$$\int_5^8\frac{2x-7}{2x+5}dx<\int_5^6 \left(\frac{8}{75}x-\frac13\right)dx+\int_6^8 \left(\frac{24}{361}x-\frac{35}{361}\right)dx\approx \\
0.2533333333+0.736842105=0.9901754383<1.$$
