Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g.  $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$    
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.

Starting from (as you almost wrote)
$$\frac{2x}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2},$$
rewrite as
$$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}
      = \frac{Ax+(2A+B)}{(x+2)^2}.$$
From this, one sees that $2x = Ax + (2A+B)$, so that $A=2$ and $2A+B=0$ and thus $B=-4$ (as you correctly derived).
An alternative method, using substitution, starts again from
$$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}$$
so that $2x = A(x+2)+B$. Substituting $x=-2$ gives $B=-4$ and thus $2x = A(x+2)-4$. Now substitute any value other than $-2$ (say $0$) for $x$, giving $0 = 2A-4$, so that $A=2$.
