Partial fractions integration Calculus II proof (1/2a)ln((a+x)/(a-x))+C Calculus II

I'm stuck on this super tough Calc II problem any help would be appreciated:
Use the method of partial fractions to verify the integration formula :

Integral{ 1/a^2-x^2 dx = (1/2a)*ln((a+x)/(a-x))+C

1/(a² - x²) = 1/(a + x)(a - x)
. . . . . . . = A/(a + x) + B/(a - x)
. . . . . . . = [A(a - x) + B(a + x)]/(a - x)(a + x)
. . . . . . . = [(A + B)a + (B - A)x]/(a² - x²)

Determining A and B
(1 + 0x)/(a² - x²) = [(A + B)a + (B - A)x]/(a² - x²)
Therefore
1 + 0x = (A + B)a + (B - A)x
Thus
1 = (A + B)a
0 = (B − A)
From there
B + A = 1/a
B − A = 0
⇒ A = B = 1/(2a)

The partial fractions of the integrand are
1/(a² - x²) = A/(a + x) + B/(a - x)
1/(a² - x²) = (1/2a)*1/(a + x) + (1/2a)*1/(a - x)

Integrating
∫ dx/(a² - x²) = (1/2a) ∫ dx/(a + x) + (1/2a) ∫ dx/(a − x)
. . . . . . . . .= (1/2a) ln|a + x| − (1/2a) ln|a − x|
. . . . . . . . .= (1/2a) ln[|a + x|/|a − x|]
. . . . . . . . .= (1/2a) ln[|a + x|/|a − x|]
. . . . . . . . .= (1/a) ln√[|a + x|/|a − x|]