Harap bantuannya tentang integral soal tertera digambar

integral dgn substitusi
∫ (4x-3) (6x -4x²)⁻²/³ dx

u = 6x - 4x²
du/dx= 6 - 8x = -2(4x -3)
4x-3  dx = -¹/₂ du

∫(4x-3) (6x-4x²)⁻²/3 dx = ∫ -¹/₂  (u)⁻²/³ du
= - ¹/₂  (3)(u)¹/³ + c
= - ³/₂  ∛u + c
= - ³/₂ ∛(6x - 4x²)