Answer
$\frac{c+1}{c-1}, with$ $c\ne1$
Work Step by Step
Given : $\frac{c^{2}+3c+2}{c^{2}-4c+3} \div \frac{c+2}{c-3}$
This becomes : $\frac{(c+1)(c+2)}{(c-1)(c-3)} \times \frac{c-3}{c+2}$
$= \frac{c+1}{c-1}$
(After dividing out the common factors (c-3) and (c+2))
