Answer
$$
\int\left(\frac{1-x}{x}\right)^{2} d x =-\frac{1}{x}-2 \ln |x|+x+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\begin{aligned}
\int\left(\frac{1-x}{x}\right)^{2} d x & =\int\left(\frac{1}{x}-1\right)^{2} d x \\
&=\int\left(\frac{1}{x^{2}}-\frac{2}{x}+1\right) d x \\
&=-\frac{1}{x}-2 \ln |x|+x+C
\end{aligned}
$$
where $C$ is an arbitrary constant.