matematika bab integral , ... tolong ya kk'

∫ 2x √(3x + 1) dx
= 2 ∫ x √(3x + 1) dx

u = 3x + 1 → x = 1/3 (u - 1)
du / dx = 3 → dx = 1/3 du
= 2 ∫ [1/3 (u - 1)] √u (1/3 du)
= 2 (1/3)(1/3) ∫ (u - 1) √u du
= 2/9 ∫ [u^(3/2) - u^(1/2)] du
= 2/9 [2/5 u^(5/2) - 2/3 u^(3/2)] + C
= 4/45 (3x + 1)^(5/2)  - 4/27 (3x + 1)^(3/2) + C