Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
 $ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this? 

$a=(x^2-1)=(x+1)(x-1)$
$b=(x^2+2x+1)=(x+1)^2=(x+1)(x+1)$
So, we have:
$$\frac{(x+3)}{a}-\frac{(x-2)}{b}$$
$$\frac{b(x+3)-a(x-2)}{ab}$$
$ab=(x+1)^3(x-1)$
$b(x+3)=(x+3)(x+1)(x+1)$
$a(x-2)=(x-2)(x+1)(x-1)$
Then,
$$\frac{(x+3)(x+1){(x+1)}-(x-2){(x+1)}(x-1)}{{(x+1)^3}(x-1)}$$
$$\frac{{(x+1)}[(x+3)(x+1)-(x-2)(x-1)]}{{(x+1)^{3}}(x-1)}$$
$$\frac{(x+3)(x+1)-(x-2)(x-1)}{(x+1)^2(x-1)}$$
Expanding:
$$\frac{7x+1}{x^3+x^2-x-1}$$
