Answer
$\int (\frac{1-x}{x})^2~dx = -\frac{1}{x}-2~ln~\vert x \vert+x+C$
Work Step by Step
$\int (\frac{1-x}{x})^2~dx$
$=\int (\frac{1-2x+x^2}{x^2})~dx$
$=\int (\frac{1}{x^2}-\frac{2x}{x^2}+\frac{x^2}{x^2})~dx$
$=\int (\frac{1}{x^2}-\frac{2}{x}+1)~dx$
$= -\frac{1}{x}-2~ln~\vert x \vert+x+C$