## Calculus 10th Edition

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