mohon bantuannya ya sahabat :))

coba dengan integral substitusi

u = x³
du = 3x²  dx
x²  dx = 1/3 du
x⁵ dx = x². x³ dx = 1/3 du. (u) = 1/3 u du
∫x⁵ √(2-3x²) dx = 1/3 ∫ u √(2-u) du
.
1/3 ∫ u√(2-u) du =
misal t = 2 - u --> u = 2 -  t
dt = - du --> du = - dt
= 1/3 ∫ (2-t) √t  (-dt)
= - 1/3 ∫ 2t¹/² - t³/²  dt
= -1/3 [ 2(2/3)(t)³/²  - (2/5)(t)⁵/²
= - 1/3 (4/3 t³/² ) - 1/3 (- 2/5)(t⁵/²)
= -4/9 t³/²  + 2/15 t⁵/²
= - 2/5 t³/²  ( 10  + 3t)
= -2/45 (2- x³)³/² ( 10 - 3(2 -x³)
= - 2/45 (2-x³)³/²  ( 10 - 6 + 3x³)
= - 2/45 (3x³+4) ( -x³ + 2)³/²  + c
...