We have to find the integral of 12/(x^2 - 4x - 12)

x^2 - 4x - 12

=> x^2 - 6x + 2x - 12

=> x(x - 6) + 2(x - 6)

=> (x + 2)(x - 6)

The factors of the denominator are (x + 2) and (x - 6.

Writing 12/(x^2 - 4x - 12) as partial fractions, use the following method.

12/(x^2 - 4x - 12) = A / (x + 2) + B/ (x -6)

12 = Ax - 6A + Bx + 2B

=> 12 = x(A + B) + 2B- 6A

A + B = 0

2B - 6A = 12

=> B - 3A = 6

=> -A - 3A = 6

=> -4A = 6

=> A = -3/2

B = 3/2

The function 12/(x^2 - 4x - 12) = (3/2)/(x - 6) - (3/2)/(x + 2)

Int [(3/2)/(x - 6) - (3/2)/(x + 2) dx]

=> Int[(3/2)/(x - 6) dx] - Int[(3/2)/(x + 2) dx] + C

=> (3/2)*ln |x - 6| - (3/2)*ln |x + 2| +C

=> (3/2)*ln |(x - 6)/(x + 2)| + C

**The required integral is (3/2)*ln |(x - 6)/(x + 2)| + C**

f(x) = 12/(x^2 -4x -12)

We need to find the integral.

First we will use partial fraction to simplify.

==> x^2 - 4x -12 = (x-6)(x+2)

==> 12/(x^2-4x-12) = A/(x-6) + A/(x+2)

Now we will multiply by x^2 -4x -12

==> 12 = A(x+2) + B(x-6).

==> 12 = (A+B)x + (2A-6B)

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

==> 2a-6B = 12

==> -8B= 12 ==> B= -12/8 = -3/2

==> A = 3/2

==> 12/(x^2 -4x-12) = 3/2(x-6) - 3/2(x+2)

Now we will find the integral.

==> Int 12/(x^2-4x-12) = (3/2)[ Int (1/(x-6) dx - Int 1/(x+2) dx ]

= (3/2)[ ln l x-6l - ln (x+2) ] + C

= (3/2) *ln (x-6)/(x+2) + C

**==> Int 12/(x^2-4x-12) = (3/2)*ln (x-6)/(x+2) + C**