partial fraction decomposition help now very quick way to solve 2x/x^2-4 as a partial fraction. I have tried the long way and it took over 30 minutes, I got it right but is it easier another way?
 A: I like to use a method that works only when you have linear terms in the denominators. 
Take $$\frac{2x}{x^2-4} = \frac{A}{x-2}+\frac{B}{x+2}$$
Now consider this:
$$A= \lim_{x \rightarrow 2} (x-2)\frac{2x}{(x-2)(x+2)}$$ and
$$B= \lim_{x \rightarrow -2} (x+2)\frac{2x}{(x-2)(x+2)}$$
Which quickly yields constants of $A=1$ and $B=1$.
EDIT: As a warning, this only really works with particularly nice problems like this one, where you only have linear terms in the denominator and the multiplicity of each term is 1. Once you start getting unfactorable terms in the denominator, you have to use the standard methods.
A: $$\frac{2x}{x^2-4}=\frac{2x}{(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2}$$
Multiplying both sides by $x^2-4$ gives
$$2x=A(x+2)+B(x-2)=(A+B)x+2(A-B)$$
Hence $A+B=2$ and $A-B=0$. The second equation gives us $A=B$ and plugging this into the first equation gives us $A=B=1$.
Hence
$$\frac{2x}{x^2-4}=\frac{1}{x-2}+\frac{1}{x+2}$$
A: Write $$\frac{2x}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$
Then note that if we multiply both sides by $x-2$ and let $x \to 2$ we obtain
Write $$\left.\frac{2x}{x+2}\right|_{x=2}=B+\left.\frac{A(x-2)}{x+2}\right|_{x=2}$$
which gives $B=1$.
Similarly, note that if we multiply both sides by $x+2$ and let $x \to -2$ we obtain
Write $$\left.\frac{2x}{x-2}\right|_{x=-2}=A+\left.\frac{B(x+2)}{x-2}\right|_{x=-2}$$
which gives $A=1$.
Putting it all together reveals that
$$\frac{2x}{x^2-4}=\frac{1}{x+2}+\frac{1}{x-2}$$
