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? 

Let's start here and simplify the problem.
$$\frac{(x+3)}{(x-1)(x+1)} - \frac{(x-2)}{(x+1)(x+1)}$$
I'm going to call $(x+1)=a$ and $(x-1)=b$
$$\frac{(x+3)}{(b)(a)} - \frac{(x-2)}{(a)(a)}$$
So, on the right side we have (a)(a) in the denominator, and only one in the denominator on the right. So we have to multiply the term on the right by $\frac{a}{a}$.
The term on the left has (b) in the denominator, while the one on the right doesn't. So we're going to to multiply the one on the right by $\frac{b}{b}$.
Let's take those steps, and we get:
$$\frac{(x+3)(a)}{(b)(a)(a)} - \frac{(x-2)(b)}{(a)(a)(b)}$$
See how both denominators are identical!? Great! Now we can put both of the numerators on top of the denominator we worked so hard to get.
$$\frac{(x+3)(a)-(x-2)(b)}{(b)(a)(a)}$$
Now all you have to do is plug back in $a=(x+1)$ and $b=(x-1)$, open up the brackets and collect like terms.
