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Math Help - a question about Integration

  1. #1
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    a question about Integration

    i can not understand how to integrate a power function like this one:
    ∫ (x² - 1) / (x³ - 3x + 1 ) ^ 6 dx
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  2. #2
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    Quote Originally Posted by mHadad View Post
    i can not understand how to integrate a power function like this one:
    ∫ (x - 1) / (x - 3x + 1 ) ^ 6 dx
    Let y = x^3 - 3x +1.
    Then dy = (3x^2 - 3)dx = 3(x^2 - 1)dx, so your integral becomes

    (1/3)*Int(dy/y^6) = (1/3)*(1/-7)1/y^7 = -1/(21y^7) = -1/[21(x^3 - 3x + 1)].

    -Dan
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  3. #3
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    why did u multiply by 1/3
    (1/3)*Int(dy/y^6)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mHadad View Post
    why did u multiply by 1/3
    (1/3)*Int(dy/y^6)
    Because dy = 3(x^2 - 1)dx and what we have in the numerator of the integrand is only (x^2 - 1)dx.

    -Dan
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  5. #5
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    u multiplied by 1/3 to get rid of 3 ?in all of the solutions to these questions the author multiplied by a constant to get rid of a number, that what i can not understand why this multiplication is done and why we should get rid of the constant (3 for example)?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mHadad View Post
    u multiplied by 1/3 to get rid of 3 ?in all of the solutions to these questions the author multiplied by a constant to get rid of a number, that what i can not understand why this multiplication is done and why we should get rid of the constant (3 for example)?
    Specifically, what I am trying to do is replace
    (x^2 - 1)dx
    with an expression in y. Since we have that
    dy = 3(x^2 - 1)dx

    (x^2 - 1)dx = (1/3)dy.

    So this gives us
    (x^2 - 1)dx/(x^3 - 3x + 1)^6 = (1/3)dy/y^6

    which we can sub in under the integration symbol.

    -Dan
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