What is partial fraction decomposition of 2x/(x^2-9)?

We'll write the fraction 2x/(x^2 - 9) as an algebraic sum
of partial fractions: [A/(x-3)] + [B/(x+3)]

Since the LCD
of the fractions from the right side is (x-3)(x+3) = x^2 - 9, we'll multiply by x^2 - 9
both fractions:

2x/(x^2 -9)= [A(x+3) + B(x-3)]/ (x^2
-9)

Having the common denominator (x^2 -9), we'll simplify
it.

2x = Ax+3A+Bx-3B

We'll
factorize by x to the right side:

2x = x*(A+B) +
(3A-3B)

Comparing, we'll
have:

A + B=2
(1)

3A-3B=0

We'll divide by
3:

A - B = 0 (2)

We'll add the
second relation to the first one:

A + B + A -
B=2+0

2A=2

A =
1

But, from (2) => A=B =
1

The partial fraction decomposition is:
2x/(x^2-9) = 1/(x-3) + 1/(x+3).