Well, I would suggest first to complete the square. You should have 9*x^2 + 6*x - 8 = 9*x^2 + 6*x + 1 - 9 = (3*x + 1)^2 - 9. Then I suggest to make a little triangle - that's how I usually did it - or just implement those little formulas you're given in your textbook. But if I make a triangle, I denote the hypotenuse as (3*x + 1), the angle U, the adjacent side = 3. so, sec(u) = (3*x + 1)/3, so (3*x+1) = 3*sec(u). You substitute your (3*x+1) back, and do not forget, that x = sec(u) - 1/3. Then dx = sec(u)*tan(ua)du. So, your new integral is (1/sqrt(9*(sec(u))^2-9))*sec(u)*tan(u)du. Try to do this!