Integral 3x (2x+1)^3 dx adalah ?

Integral Substitusi.

∫ 3x(2x + 1)³ dx = 3 ∫ x(2x + 1)³ dx
u = 2x + 1 → x = (u - 1) / 2
dx = du / 2
= 3 ∫ [(u - 1) / 2] (du / 2)
= 3/4 ∫ (u - 1)u³ du
= 3/4 ∫ (u⁴ - u³) du
= 3/4 (1/5 u⁵ - 1/4 u⁴) + C
= 3/20 (2x + 1)⁵ - 3/16 (2x + 1)⁴ + C
= (2x + 1)⁴ [-3/16 + 3/20 (2x + 1)] + C
= (2x + 1)⁴ (3/10 x - 3/80) + C
= 3/80 (2x + 1)⁴ (8x - 1) + C