Answer
$\displaystyle \frac{x-1}{x+2},\ \qquad x\neq-2,-1$
Work Step by Step
Factor each denominator.
$x^{2}+3x+2= (x+1)(x+2)$
... two factors of $2$ with sum $3$ are $+1$ and $+2.$
$LCD=(x+1)(x+2)$
$\displaystyle \frac{4x^{2}+x-6}{x^{2}+3x+2}-\frac{3x}{x+1}+\frac{5}{x+2}= \displaystyle \frac{4x^{2}+x-6}{(x+1)(x+2)}-\frac{3x}{x+1}\times\frac{x+2}{x+2}+\frac{5}{x+2}\times\frac{x+1}{x+1}$
$= \displaystyle \frac{4x^{2}+x-6}{(x+1)(x+2)}- \displaystyle \frac{3x^{2}+6x}{(x+1)(x+2)}+ \displaystyle \frac{5x+5}{(x+1)(x+2)}$
$=\displaystyle \frac{4x^{2}+x-6-3x^{2}-6x+5x+5}{(x+1)(x+2)}$
$=\displaystyle \frac{x^{2}-1}{(x+1)(x+2)}\qquad$... recognize a difference of squares
$=\displaystyle \frac{(x-1)(x+1)}{(x+1)(x+2)}\qquad$... We have a common factor. Reduce.
$=\displaystyle \frac{x-1}{x+2},\ \qquad x\neq-2,-1$
(-1 is also excluded as $(x+1)$ was a part of the expression before reducing)
