## Calculus 8th Edition

$2\frac{7}{24}$
First, we expand the terms being squared and get the polynomial $4 - \frac{4}{p^2} + \frac{1}{p^4}$ as our integrand. We can rewrite the terms to get rid of the fractions and get $4 - 4p^{-2} + {p^{-4}}$ Evaluating this integral gives us $4p - (-)4p^{-1} +\frac{p^{-3}}{-3} = 4p + \frac{4}{p} -\frac{1}{3p^3}$ We now substitute our bounds of integration $2$ and $1$ and get $(4(2) + \frac{4}{2} - \frac{1}{3(2)^3}) - (4 + 4 - \frac{1}{3}) = (8 + 2 - \frac{1}{24}) - (8 - \frac{1}{3}) = 2 - \frac{1}{24} + \frac{8}{24} = 2\frac{7}{24}$