Answer
$\frac{3x+2}{(x+2)(x-1)}$.
Work Step by Step
The given expression is
$\frac{x+1}{x^2+x-2}-\frac{1}{x^2-3x+2}+\frac{2x}{x^2-4}$
Factor each denominator.
$\Rightarrow x^2+x-2$
Rewrite the middle term $x$ as $2x-x$.
$\Rightarrow x^2+2x-x-2$
Group terms.
$\Rightarrow (x^2+2x)+(-x-2)$
Factor each group.
$\Rightarrow x(x+2)-1(x+2)$
Factor out $(x+2)$.
$\Rightarrow (x+2)(x-1)$
$\Rightarrow x^2-3+2$
Rewrite the middle term $-3x$ as $-2x-x$.
$\Rightarrow x^2-2x-x+2$
Group terms.
$\Rightarrow (x^2-2x)+(-x+2)$
Factor each group.
$\Rightarrow x(x-2)-1(x-2)$
Factor out $(x-2)$.
$\Rightarrow (x-2)(x-1)$
$\Rightarrow x^2-4$
$\Rightarrow x^2-2^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (x+2)(x-2)$
Back substitute all factors into the fraction.
$\Rightarrow \frac{x+1}{(x+2)(x-1)}-\frac{1}{(x-2)(x-1)}+\frac{2x}{(x+2)(x-2)}$
The LCD of the denominators is $(x+2)(x-2)(x-1)$
$\Rightarrow \frac{(x+1)(x-2)}{(x+2)(x-2)(x-1)}-\frac{(x+2)}{(x+2)(x-2)(x-1)}+\frac{2x(x-1)}{(x+2)(x-2)(x-1)}$
$\Rightarrow \frac{(x+1)(x-2)-(x+2)+2x(x-1)}{(x+2)(x-2)(x-1)}$
Use FOIL method and the distributive property to clear parentheses.
$\Rightarrow \frac{x^2-2x+x-2-x-2+2x^2-2x}{(x+2)(x-2)(x-1)}$
Simplify.
$\Rightarrow \frac{3x^2-4x-4}{(x+2)(x-2)(x-1)}$
Factor the numerator.
$\Rightarrow 3x^2-4x-4$
Rewrite the middle term $-4x$ as $-6x+2x$.
$\Rightarrow 3x^2-6x+2x-4$
Group terms.
$\Rightarrow (3x^2-6x)+(2x-4)$
Factor each group.
$\Rightarrow 3x(x-2)+2(x-2)$
Factor out $(x-2)$.
$\Rightarrow (x-2)(3x+2)$
Back substitute into the fraction.
$\Rightarrow \frac{(x-2)(3x+2)}{(x+2)(x-2)(x-1)}$
Cancel common terms.
$\Rightarrow \frac{(3x+2)}{(x+2)(x-1)}$.
