minta tolong ya buat besok UAS

Intergral substitusi

1. ∫(x²+2) √(x³+6x +1) dx
u = x³+6x + 1
du = 3x² +6 dx = 3(x²+2) dx
(x² + 2) dx = 1/3 du

∫(x+2) √(x³+6x +1) dx = ∫ 1/3 U¹/² du = 1/3 (2/3) u³/² = 2/9 u√ u
∫(x+2) √(x³+6x+1) dx = ²/₉ (x³ +6x +1) √(x³ + 6x + 1) + c

2. ∫(3 - x) / √(x²-6x+5) dx
u = x² - 6x + 5
du = 2x - 6 dx = -2(3 - x) dx
(3-x) dx = - 1/2 du
∫ - 1/2 u ⁻¹/²  du  = -1/2 (1/2) u¹/² + c = -1/4 √u + c
∫(3-x)/√(x²-6x+5) dx = -1/4 √(x²-6x+5) + c