Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 70

Find the integral $\displaystyle \int^2_1 \frac{4+u^2}{u^3} du$

$\begin{equation} \begin{aligned} \int^2_1 \frac{4+u^2}{u^3} du &= \int^2_1 \frac{4}{u^3} + \frac{u^2}{u^3} du\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \int^2_1 4u^{-3} + \frac{1}{u} du\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \left[4 \left( \frac{u^{-3+1}}{-3+1}\right) + \ln u \right]^2_1\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \left[4 \left( \frac{u^{-2}}{-2}\right) + \ln u \right]^2_1\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \left[ -2u^{-2} + \ln u \right]^2_1\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \frac{-2}{(2)^2} + \ln (2) - \left[ \frac{-2}{(-1)^2} + \ln (1) \right]\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \frac{-2}{4} + \ln (2) + 2 - 0\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \frac{-1}{2} + 2 + \ln (2)\\ \\ \int^2_1 \frac{4+u^2}{u^3} du &= \frac{3}{2} + \ln (2) \end{aligned} \end{equation}$