Answer
$y=x-2\ln|x|+1$
When C=1, the graph passes through $(-1,0)$
Work Step by Step
$\displaystyle \frac{x-2}{x}=\frac{x}{x}-\frac{2}{x}=1-\frac{2}{x}$,
$y=\displaystyle \int(1-\frac{2}{x})dx=\int dx-2\int\frac{dx}{x}$
$y=x-2\ln|x|+C$
$(x,y)=(-1, 0)$ is on the graph, so we find C
$0=-1-2\ln|-1|+C$
$0=-1+0+C$
$C=1$
$y=x-2\ln|x|+1$
