hasil integral 2x(3x-2)^3 dx=

2x(3x-2)^3 = 2x[(3x-2)(3x-2)(3x-2)]
= 2x[(6x^2-6x-6x+4)(3x-2)]
= 2x[(6x^2-12x+4)(3x-2)]
= 2x(18x^3-12x^2-36x^2+24x-4)
= 2x(18x^3-48x^2+24x-4)
= 36x^4-96x^3+48x^2-8x
§ x dx = x+c
§2x(3x-2)^2 dx = 26x^4-96x^3+48x^2-8x+c