## Calculus 10th Edition

$y=2\ln|x-9|+C$ When C=$4-2\ln 9$, the graph passes through $(4,0)$
$\displaystyle \frac{2x}{x^{2}-9x}=\frac{2x}{x(x-9)}=\frac{2}{x-9 }, x\neq 0$ $y=\displaystyle \int\frac{2}{x-9 }dx=2\int\frac{1}{x-9 }dx\qquad\left[\begin{array}{l} u=x-9\\ du=dx \end{array}\right]$ $=2\displaystyle \int\frac{1}{u}du$ $=2\ln|u|+C$ $y=2\ln|x-9|+C$ $(x,y)=(0, 4)$ is on the graph, so we find C $4=2\ln|0-9|+C$ $C=4-2\ln 9$