Solve the indefinite integral of the function f(x)=1/(9x^2-6x-8)?

We notice that we can create a perfect square to the
denominator of the integrand.

9x^2 - 6x - 8 = 9x^2 - 6x + 1
- 9 = (3x - 1)^2 - 9

We'll substitute the binomial 3x - 1 =
t.

We'll differentiate both
side:

3dx = dt => dx =
dt/3

We'll re-write the integral in
t:

Int dx/[(3x - 1)^2 - 9] = Int dt/3(t^2 -
9)

Int dt/3(t^2 - 9) = (1/18)*ln|(t-3)/(t+3)| +
C

The indefinite integral of the function
f(x) is: Int f(x)dx = (1/18)*ln|(3x-4)/(3x+2)| +
C