integral
substitusi
∫ (6x² +4x) (√(x³ + x² -7) dx
u = x³ + x²- 7
du = (3x² +2x) dx
untuk (6x² + 4x) dx = 2(3x² +2x) dx = 2 du
∫ (2 du) (√u) = 2 ∫u^(1/2) du
= 2 (2/3) (u)^(3/2) + c
= (4/3) u√ u + c
atau
= (4/3) (x³ + x² - 7) √(x³ + x² - 7) + c
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Verified answer
integral
substitusi
∫ (6x² +4x) (√(x³ + x² -7) dx
u = x³ + x²- 7
du = (3x² +2x) dx
untuk (6x² + 4x) dx = 2(3x² +2x) dx = 2 du
∫ (2 du) (√u) = 2 ∫u^(1/2) du
= 2 (2/3) (u)^(3/2) + c
= (4/3) u√ u + c
atau
= (4/3) (x³ + x² - 7) √(x³ + x² - 7) + c