using method of partial fractions to decompose this 1/(x-3)(x+3)

Question

is 1/6(x-3)-1/6(x+3) correct or have i gone wrong somewhere

Answer 1

I get

1/(x-3)(x+3) = A/(x-3) + B/(x+3)

1 = (x+3)A + (x-3)B

1 = x(A+B) + 3A -3B

-> A + B = 0 -> A = -B

-> 3A - 3B = 1 -> A = 1/6; B = -1/6

= (1/6)(x-3) - (1/6)/(x+3)

Answer 2

thanks so it is the same thank god for that