Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 64

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


$\begin{equation} \begin{aligned} \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \int^2_1 \left( 4u^{-3} + \frac{1}{u} \right) du \\ \\ \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \int^2_1 4u^{-3} du + \int^2_1 \frac{du}{u} \\ \\ \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \left[ \frac{4u^{-2}}{-2} \right]^2_1 + [\ln u]^2_1 \\ \\ \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \left[ \frac{-2}{u^2} \right]^2_1 + [\ln u]^2_1 \\ \\ \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \left[ \frac{-2}{2^2} - \left( \frac{-2}{12} \right) \right] + [\ln 2 - \ln 1] \\ \\ \int^2_1 \left( \frac{4 + u^2}{u^3} \right) du =& \frac{3}{2} + \ln 2 \end{aligned} \end{equation}$