Metode substitusi 1. ∫ x² / √x³-5 dx = ....... 2. ∫ (9x² - 6x) √x³-x²+3 dx = .......

1) ∫ x² / √(x³ - 5) dx
= ∫ x² (x³ - 5)^(-1/2) dx

Misal u = x³ - 5
=> du = 3x² dx
=> dx = du/(3x²)

= ∫ x² (x³ - 5)^(-1/2) dx
= ∫ x² u^(-1/2) du/(3x²)
= ∫ 1/3 u^(-1/2) du
= (1/3)/(1/2) u^(1/2) + C
= 2/3 √u + C
= 2/3 √(x³ - 5) + C

2) ∫ (9x² - 6x) √(x³ - x² + 3) dx
= ∫ 3(3x² - 2x) (x³ - x² + 3)^(1/2) dx

Misal : u = x³ - x² + 3
=> du = 3x² - 2x dx
=> dx = du/(3x² - 2x)

= ∫ 3(3x² - 2x) (x³ - x² + 3)^(1/2) dx
= ∫ 3(3x² - 2x) u^(1/2) du/(3x² - 2x)
= ∫ 3u^(1/2) du
= 3/(3/2) u^(3/2) + C
= 2 u√u + C
= 2 (x³ - x² + 3)√(x³ - x² + 3) + C

^ = pangkat