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Math Help - Did I solve this Integral Correctly?

  1. #1
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    Did I solve this Integral Correctly?

    X^2 / third root X+3

    u = X+3
    du = dx
    x= u-3

    (u-3)^2 + (u)^-1/3

    (u-3)^3 / 3 + (3/2) (u)^2/3

    1/3(x+3)^3 + 3/2(x+3)^2/3 + c
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  2. #2
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    Quote Originally Posted by gammaman View Post
    X^2 / third root X+3

    u = X+3
    du = dx
    x= u-3

    (u-3)^2 + (u)^-1/3

    (u-3)^3 / 3 + (3/2) (u)^2/3

    1/3(x+3)^3 + 3/2(x+3)^2/3 + c

    Your substitution isn't quite right. First, yes, if we let
    u = X+3
    du = dx

    But the next part is highly flawed. Though true that
    x = u-3

    We're interested in what x^2 is, because that's the numerator. You need to square both sides. Also, when trying to turn this into a product, you need to multiply the numerator by the inverse of the denominator, not add it.
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  3. #3
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    =You need to square both sides. Also, when trying to turn this into a product, you need to multiply the numerator by the inverse of the denominator, not add it.
    Can you show me?
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