1. ∫x(x+4)^5 dx 2. ∫4x(4x^2-3)^4 dx

coba dgn integral substitusi

∫ x(x+4)⁵ dx
u = x+ 4
du/dx = 1 --> dx = du
.
u = x + 4 --> x= u - 4
.
∫ x (x+4)⁵ dx = ∫ (u-4)(u)⁵ du
= ∫ u⁶ - 4u⁵ du
= ¹/₇ u⁷ - ⁴/₆ u⁶ + c
= ¹/₂₁ U⁶ ( 3 u - 14) + c
= ¹/₂₁ (x+4)⁶ {3(x+4) - 14} + c
= ¹/₂₁ (x+4)⁶ (3x +12 -14) + c
= ¹/₂₁ (x+4)⁶ (3x-2)+ c
= ¹/₂₁ (3x-2)(x+4)⁶ + c
..

2) ∫4x (4x² -3)⁴ dx =..
u = 4x² - 3
du/dx = 8x --> 4x dx = 2 du
∫ 4x (4x² -3)⁴ dx = ∫ 2 u⁴ du = ²/₅ u⁵ + c
∫ 4x (4x²-3)⁴ dx = ²/₅ (4x² -3)⁵ + c