integral parsial 24x(3x-5)pangkat6 dx

saya bisa pakai substitusi tapi
misalkan 3x - 5 = u
3 dx = du
dx = du/3
selain itu
3x - 5 = u
3x = u + 5
x = (u + 5)/3
ubah dalam u persamaan nya
(24(u+5)/3 u^6) du/3
= (8(u+5) u^6) du/3
= ⅓(8u + 40)u^6 du
= ⅓(8u^7 + 40u^6) du
baru integralkan
= ⅓ {(8/(7+1)u^8 + (40/(6+1)u^7)} + C
= ⅓ {u^8 + (40/7)u^7} + C
= (⅓)(1/7) {7u^8 + 40u^7} + C
= (1/21)(u^7)(7u + 40) + C
ingat u = 3x - 5
= (1/21)(3x - 5)^7 (7(3x-5) + 40) + C
= (1/21)(3x - 5)^7 (21x - 35 + 40) + C
= (1/21) (3x - 5)^7 (21x + 5) + C
= (1/21) (21x + 5) (3x - 5)^7 + C
keterangan ^ tanda pangkat