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?

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
$$\frac{2x}{x^2-4}=\frac{1}{x+2}+\frac{1}{x-2}$$
