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