You need to use the following formula that helps you to directly integrate, such that:

`int 1/(x^2 - a^2) dx = (1/(2a))ln|(x - a)/(x + a)| + c`

Reasoning by analogy, yields:

`int 1/(x^2 - 4)dx = int 1/(x^2 - 2^2)dx`

`int 1/(x^2 - 2^2)dx = 1/(2*2)ln|(x - 2)/(x + 2)| + c`

**Hence, directly integrating, using the indicated formula, yields **`int 1/(x^2 - 2^2)dx = ln root(4)(|(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 ofÂ 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

BÂ = -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