#### Answer

The solution is $x=3$

#### Work Step by Step

$\dfrac{3}{x^{2}+x-2}-\dfrac{1}{x^{2}-1}=\dfrac{7}{2x^{2}+6x+4}$
Factor all rational expressions completely:
$\dfrac{3}{(x+2)(x-1)}-\dfrac{1}{(x-1)(x+1)}=\dfrac{7}{2(x+2)(x+1)}$
Multiply the whole equation by $2(x+2)(x-1)(x+1)$:
$2(x+2)(x-1)(x+1)\Big[\dfrac{3}{(x+2)(x-1)}-\dfrac{1}{(x-1)(x+1)}=\dfrac{7}{2(x+2)(x+1)}\Big]$
$3(2)(x+1)-(2)(x+2)=7(x-1)$
Evaluate the indicated operations:
$6x+6-2x-4=7x-7$
Take all terms with $x$ to the left side and all terms without $x$ to the right side:
$6x-2x-7x=-7-6+4$
Simplify both sides:
$-3x=-9$
Solve for $x$:
$x=\dfrac{-9}{-3}$
$x=3$