Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 70

Find the integral $\displaystyle \int^2_1 \frac{(x - 1)^3}{x^2} dx$


$\begin{equation} \begin{aligned} & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \int^2_1 \frac{x^3 - 3x^2 + 3x - 1}{x^2} dx \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \int^2_1 \left( \frac{x^3}{x^2} - \frac{3x^2}{x^2} + \frac{3x}{x^2} - \frac{1}{x^2} \right) dx \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \int^2_1 \left( x - 3 + \frac{3}{x} - \frac{1}{x^2} \right) dx \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \int^2_1 \left( x - 3 - \frac{3}{x} - x^{-2} \right) dx \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \left[ \frac{x^2}{2} - 3x + 3|n|x| + x^{-1} \right]^2_1 \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \frac{(2)^2}{2} - 3(2) + 3 ln 2 + \frac{1}{2} - \left[ \frac{(1)^2}{2} - 3 (1) + 3 ln 1 + \frac{1}{1} \right] \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = 2 - 6 + 3ln 2 + \frac{1}{2} - \frac{1}{2} + 3 - 3(0) -1 \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = 3ln 2 - 2 \\ \\ & \int^2_1 \frac{(x - 1)^3}{x^2} dx = \int^2_1 \end{aligned} \end{equation}$