How do you integrate this?

9x^2 - 12x + 8 = (3x-2)^2 + 2^2
2x + 3 = (2/3)(3x-2) + 13/3

Let 3x-2 = 2 tanθ
3 dx = 2 sec^2 θ dθ
dx = (2/3) sec^2 θ dθ

2x + 3 = (2/3)(3x-2) + 13/3 = (4/3)tanθ + 13/3
9x^2 - 12x + 8 = (3x-2)^2 + 2^2 = 4 tan^2 θ + 4 = 4 sec^2 θ

∫ [ (2x+3)/(9x^2-12x+8) ] dx
= ∫ [ (4/3)tanθ + 13/3 ] * [ (2/3) sec^2 θ dθ] / [ 4 sec^2 θ ]
= (1/6) * ∫ [ (4/3)tanθ + 13/3 ] dθ
= (2/9) * ∫ tanθ dθ + (13/18)θ
= (2/9) * ln ∣ secθ ∣ + (13/18)θ + C
= (2/9) * ln ∣ [ √(9x^2-12x+8) ] / 2 ∣ + (13/18) * tan^ -1 [ (3x-2) / 2 ] + C

Ans: (2/9) * ln ∣ [ √(9x^2-12x+8) ] / 2 ∣ + (13/18) * tan^ -1 [ (3x-2) / 2 ] + C

Note:
3x-2 = 2 tanθ
tan θ = (3x-2) / 2
θ = tan^ -1 [ (3x-2) / 2 ]

tan θ = (3x-2) / 2
sec θ = √[ (3x-2)^2 + 2^2 ] / 2 = [ √(9x^2-12x+8) ] / 2