Hasil dari integral 2x (6x-1)^(1/3) dx adalah ....

(8x + 1)/56 * (6x - 1)^(4/3) + C

Pembahasan

Untuk menyelesaikan integral dari 2x(6x-1)^(1/3) dx, kita dapat menggunakan metode substitusi. Misalkan u = 6x - 1. Maka, du/dx = 6, sehingga dx = du/6.

Kita juga perlu mengekspresikan x dalam bentuk u. Dari u = 6x - 1, kita dapatkan 6x = u + 1, sehingga x = (u + 1)/6.

Sekarang kita substitusikan ke dalam integral:
∫ 2x (6x-1)^(1/3) dx = ∫ 2 * ((u + 1)/6) * u^(1/3) * (du/6)

Sederhanakan konstanta:
= ∫ (2/36) * (u + 1) * u^(1/3) du
= (1/18) ∫ (u * u^(1/3) + 1 * u^(1/3)) du
= (1/18) ∫ (u^(4/3) + u^(1/3)) du

Sekarang, integralkan terhadap u:
= (1/18) * [ (u^(4/3 + 1))/(4/3 + 1) + (u^(1/3 + 1))/(1/3 + 1) ] + C
= (1/18) * [ (u^(7/3))/(7/3) + (u^(4/3))/(4/3) ] + C
= (1/18) * [ (3/7)u^(7/3) + (3/4)u^(4/3) ] + C

Masukkan kembali u = 6x - 1:
= (1/18) * [ (3/7)(6x - 1)^(7/3) + (3/4)(6x - 1)^(4/3) ] + C

Faktorkan keluar (6x - 1)^(4/3) dan sederhanakan konstanta:
= (1/18) * (6x - 1)^(4/3) * [ (3/7)(6x - 1) + 3/4 ] + C
= (1/18) * (6x - 1)^(4/3) * [ (18x - 3)/7 + 3/4 ] + C
= (1/18) * (6x - 1)^(4/3) * [ (4(18x - 3) + 7(3)) / 28 ] + C
= (1/18) * (6x - 1)^(4/3) * [ (72x - 12 + 21) / 28 ] + C
= (1/18) * (6x - 1)^(4/3) * [ (72x + 9) / 28 ] + C
= (9(8x + 1)) / (18 * 28) * (6x - 1)^(4/3) + C
= (8x + 1) / (2 * 28) * (6x - 1)^(4/3) + C
= (8x + 1) / 56 * (6x - 1)^(4/3) + C

Jadi, hasil dari integral 2x (6x-1)^(1/3) dx adalah (8x + 1)/56 * (6x - 1)^(4/3) + C.