Answer
$y=-3\ln|x-2|+C$
when $C=0$, the graph passes through (1,0)
Work Step by Step
$y=\displaystyle \int\frac{3}{2-x}dx =-3\displaystyle \int\frac{1}{x-2}dx$
$\left[\begin{array}{l}
u=x-2\\
du=dx
\end{array}\right]$
$=-3\displaystyle \int\frac{1}{u}du$
$=-3\ln|u|+C$
$y=-3\ln|x-2|+C$
$(x,y)=(1, 0)$ is on the graph, so we find C
$0=-3\ln|1-2|+C$
$0=-3\ln 1+C$
$C=0$
$y=-3\ln|x-2|$
