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

### 2 Answers | Add Yours

Let f(x)= 1/(x^2 -4)

We need to determine F(x) such that:

F(x) = intg f(x) dx

==>F(x) = intg [ 1/(x^2 - 4)] dx

Let us simplify:

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

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

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

Open brackets:

==> Ax 2A + Bx - 2B = 1

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

==> A+b = 0 .....(1)

==> A-b = 1/2......(2)

Now add (1) and (2):

==> 2A = 1/2

==>** A = 1/4**

**==> B = -1/4**

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

==> F(x) = intg [1/4(x-2) - 1/4(x+2) ] dx

= (1/4)*( intg 1/(x-2) dx - intg 1/(x+2) dx

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

= (1/4) ln (x-2)/(x+2) + C

**==> F(x) = (1/4)*ln (x-2)/(x+2) + C **

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**