∫x^2 squareroot x^3+5 dx=?
回答 (2)
2016-07-12 8:11 am
Let u = x^3 + 5
Then, du = 3x^2 dx
∫x^2 √(x^3 + 5) dx
= ∫√(x^3 + 5) (x^2 dx)
= (1/3) ∫√(x^3 + 5) (3x^2 dx)
= (1/3) ∫√u du
= (1/3) * (2/3) u^(3/2) + C
= (2/9)(x^3 + 5)^(3/2) + C
Then, du = 3x^2 dx
∫x^2 √(x^3 + 5) dx
= ∫√(x^3 + 5) (x^2 dx)
= (1/3) ∫√(x^3 + 5) (3x^2 dx)
= (1/3) ∫√u du
= (1/3) * (2/3) u^(3/2) + C
= (2/9)(x^3 + 5)^(3/2) + C
2016-07-12 8:11 am
∫x²√(x³+5) dx
Let u = x³+5. Then du = 3x²dx, and (1/3)du = x²dx. Substitute, giving
∫√u·(1/3)du, or (1/3)∫√u du.
This evaluates to (1/3)·(2/3)u^(3/2) + C, or (2/9)u^(3/2) + C.
Substituting x³+5 for u, we get
F(x) = (2/9)(x³+5)^(3/2) + C.
Let u = x³+5. Then du = 3x²dx, and (1/3)du = x²dx. Substitute, giving
∫√u·(1/3)du, or (1/3)∫√u du.
This evaluates to (1/3)·(2/3)u^(3/2) + C, or (2/9)u^(3/2) + C.
Substituting x³+5 for u, we get
F(x) = (2/9)(x³+5)^(3/2) + C.
收錄日期: 2021-05-01 14:08:43
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