Find the integral of f(x)= 12/(x^2-4x-12)

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