beserta caranya kalau bisa

No. 18
Misal
u = 1 - 2x²
du = -4x dx
-du/4 = x dx

Jadi integral dapat ditulis
-3/4 ∫ √u  du
= -3/4 (2/3) u^(3/2) + c
= (-1/2) u^(3/2) + c
= (-1/2) (1-2x²)^(3/2) + c   ==>  D

Nomor 19
Misal 
u = sin x
du = cos x dx
Maka ∫ sin⁴x  cos x  dx
= ∫ u⁴ du
= u⁵/5 + c
= (1/5)sin⁵x + c   ==> A

Nomor 20
Misal
u = x³ + 1
du = 3x² dx
du/3 = x² dx
Jangan lupa mengubah batas integral
u = x³ + 1
u = 1 + 1 = 2
u = 0 + 1 = 1

Maka ∫ 2x² (x³ + 1)⁵ dx  {0...1}
= 2/3 ∫ u⁵ du  { 1...2}
= 2/3 [u⁶/6]  {1....2}
= 1/9 [u⁶]  {2...1}
= 1/9 [2⁶ - 1⁶]
= 1/9 [64 - 1]
= 63/9
= 7