Answer
$\frac{x^2+5x+8}{(x+2)(x+1)}$.
Work Step by Step
The given expression is
$=\left ( \frac{2x+3}{x+1}\cdot \frac{x^2+4x-5}{2x^2+x-3}\right )-\frac{2}{x+2}$
Factor all terms in the bracket
$=x^2+4x-5$
Rewrite the middle term $4x$ as $5x-x$
$=x^2+5x-x-5$
Group terms.
$=(x^2+5x)+(-x-5)$
Factor each term.
$=x(x+5)-1(x+5)$
Factor out $(x+5)$.
$=(x+5)(x-1)$
$=2x^2+x-3$
Rewrite the middle term $x$ as $3x-2x$
$=2x^2+3x-2x-3$
Group terms.
$=(2x^2+3x)+(-2x-3)$
Factor each term.
$=x(2x+3)-1(2x+3)$
Factor out $(2x+3)$.
$=(2x+3)(x-1)$
Substitute all factors into the given expression.
$=\left ( \frac{2x+3}{x+1}\cdot \frac{(x+5)(x-1)}{(2x+3)(x-1)}\right )-\frac{2}{x+2}$
Cancel common terms.
$= \frac{x+5}{x+1}-\frac{2}{x+2}$
$=(x+2)(x+1)$.
Multiply all fractions to make denominators equal.
$= \frac{x+5}{x+1}\times \frac{(x+2)}{(x+2)}-\frac{2}{x+2}\times \frac{(x+1)}{(x+1)}$
Simplify.
$= \frac{(x+5)(x+2)}{(x+2)(x+1)}-\frac{2(x+1)}{(x+2)(x+1)}$
$= \frac{(x+5)(x+2)-2(x+1)}{(x+2)(x+1)}$
$= \frac{x^2+5x+2x+10-2x-2}{(x+2)(x+1)}$
$= \frac{x^2+5x+8}{(x+2)(x+1)}$.
