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

First wewill implify using partial fractions:

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

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

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

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

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

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

==>2(A-B)= 1

==>2(-B-B) = 1

==> -2B = 1/2

==>B = -1/4

==>A = 1/4

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

Now let us integrate:

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

= (1/4) intg (1/(x-2) - 1/(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

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

To integrate f(x) = 1/(x^2-4).

We know that 1/(x^2-4) = 1/(x-2)(x+2).

1/(x^2-4) = (1/4){1/(x-2) - 1(x+2)}.

Therefore Int dx/(x^2-4) = (1/4) Int {1/(x-2) - 1/(x-2)} dx

Int dx/(x^2-4) = (1/4 ) log(x-4) - log(x+4)

Int dx/(x^2-4) = (1/4) log(x-4)/(x+4) +Constant.

We'll apply the formula of integrating a function:

f(x) = 1/(x^2 - a^2)

Int f(x)dx = Int dx/(x^2 - a^2)

Int dx/(x^2 - a^2) = (1/2a)*ln |(x-a)/(x+a)| + C

We recognize the denominator of the given function as being a difference of squares:

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

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

Int f(x)dx = Int dx/(x^2 - 2^2)

We'll substitute a by 2 and we'll get:

Int dx/(x^2 - 2^2) = (1/2*2)*ln |(x-2)/(x+2)| + C

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