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

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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**