Answer
$$3\ln \left(2\right)-2$$
Work Step by Step
Given
$$\int_{1}^{2} \frac{(x-1)^{3}}{x^{2}} d x$$
Since
\begin{aligned}
\int_{1}^{2} \frac{(x-1)^{3}}{x^{2}} d x&= \int_{1}^{2} \frac{(x-1)(x-1)^{2}}{x^{2}} d x\\
&= \int_{1}^{2} \frac{(x-1)(x^2-2x+1)}{x^{2}} d x\\
&= \int_1^2\frac{ x^3-3x^2+3x-1}{x^2}dx\\
&= \int_1^2(x-3+\frac{1}{x}-x^{-2})dx\\
&= \frac{1}{2}x^2-3x+\ln (x) +\frac{1}{x}\bigg|_1^2\\
&= \frac{3}{2}-3+3\ln \left(2\right)-\frac{1}{2}\\
&= 3\ln \left(2\right)-2
\end{aligned}
