Answer
$\frac{1}{3} \leq \int_{3}^{5} \frac{1}{x+1}~dx \leq \frac{1}{2}$
Work Step by Step
On the interval $~~3 \leq x \leq 5$:
$\frac{1}{6} \leq \frac{1}{x+1} \leq \frac{1}{4}$
Therefore, by Property 8:
$\frac{1}{6}(5-3) \leq \int_{3}^{5} \frac{1}{x+1}~dx \leq \frac{1}{4}(5-3)$
$\frac{1}{3} \leq \int_{3}^{5} \frac{1}{x+1}~dx \leq \frac{1}{2}$
