Answer
$\frac{x^{2}+5x+8}{(x+1)(x+2)}; x \ne -1,-2,-\frac{3}{2};$
Work Step by Step
$ (\frac{2x+3}{x+1} . \frac{x^{2}+4x-5}{2x^{2}+x-3}) - \frac{2}{x+2}$
Factors of $x^{2}+4x-5$ are $(x-1)(x+5)$ and factors of $2x^{2}+x-3$ are $(x-1)(2x+3)$
$ =(\frac{2x+3}{x+1} . \frac{(x-1)(x+5)}{(x-1)(2x+3)}) - \frac{2}{x+2}; x \ne -1,-2,-\frac{3}{2};$
Cross out common factors.
$ =\frac{x+5}{x+1} - \frac{2}{x+2}; x \ne -1,-2,-\frac{3}{2};$
Taking LCD,
$ =\frac{(x+5)(x+2) - 2(x+1)}{(x+1)(x+2)} ; x \ne -1,-2,-\frac{3}{2};$
$ =\frac{x^{2}+5x+2x+10 - 2(x+1)}{(x+1)(x+2)} ; x \ne -1,-2,-\frac{3}{2};$
$ =\frac{x^{2}+5x+2x+10 - 2x-2}{(x+1)(x+2)} ; x \ne -1,-2,-\frac{3}{2};$
$ =\frac{x^{2}+5x+8 }{(x+1)(x+2)} ; x \ne -1,-2,-\frac{3}{2};$
