What is the indefinite integral of y = 1/( x^2 - 4 ) ?

We notice that the denominator of the function is a
difference of squares.

We'll re-write the function as a sum
of elementary fractions:

1/(x^2-4) =
1/(x-2)(x+2)

1/(x-2)(x+2) = A/(x-2) +
B/(x+2)

We'll multiply the first ratio from the right side,
by (x+2), and the second ratio, by (x-2).

1 = A(x+2) +
B(x-2)

We'll remove the brackets from the right
side:

1 = Ax + 2A + Bx -
2B

We'll combine the like
terms:

1 = x(A+B) + 2(A-B)

For
the equality to hold, the like terms from both sides have to be
equal:

A+B = 0

A =
-B

2(A-B) = 1

We'll divide by
2:

A-B = 1/2

A+A =
1/2

2A = 1/2

We'll divide by
2:

A = 1/4

B =
-1/4

The function 1/(x^2 - 4) = 1/4(x-2) -
1/4(x+2)

Int dx/(x^2 - 4) = (1/4)*[Int dx/(x-2) -
Intdx/(x+2)]

We'll solve Int dx/(x-2) using substitution
technique:

We'll note (x-2) =
t

We'll differentiate both
sides:

dx = dt

 Int dx/(x-2) =
Int dt/t

Int dt/t = ln t + C = ln (x-2) +
C

Intdx/(x+2) = ln (x+2) +
C

Int dx/(x^2 - 4) = (1/4)*[ln (x-2)-ln (x+2)] +
C

 We'll use the quotient property of the
logarithms:

Int dx/(x^2 - 4) = (1/4)*[ln
(x-2)/(x+2)] + C