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Math Help - Indefinite integrals

  1. #1
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    Exclamation Indefinite integrals

    Question:
     \int(x^2-4x+4)^\frac{4}{3} . dx

    i know that x^2-4x+4 simplifies to  (x-2)^2
    So you can say the equation is:
     \int (x-2)^\frac{8}{3} .dx

    But i dont know how to solve it.
    Please answer in steps Thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int(x^2-4x+4)^\frac{4}{3} . dx

    i know that x^2-4x+4 simplifies to  (x-2)^2
    So you can say the equation is:
     \int (x-2)^\frac{8}{3} .dx

    But i dont know how to solve it.
    Please answer in steps Thank you
    Let u = x-2 \Rightarrow dx = du.

    Then: \int (x-2)^\frac{8}{3} ~dx = \int u^{\frac{8}{3}} ~ du

    Can you solve it now?
    Last edited by Defunkt; January 9th 2010 at 06:24 AM. Reason: typo
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  3. #3
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    Quote Originally Posted by mj.alawami View Post
    Question:
     \int(x^2-4x+4)^\frac{4}{3} . dx

    i know that x^2-4x+4 simplifies to  (x-2)^2
    So you can say the equation is:
     \int (x-2)^\frac{8}{3} .dx

    But i dont know how to solve it.
    Please answer in steps Thank you
    You are DEFINITELY on the right track.

    Now use a u substitution.

    Let u = x - 2 so that \frac{du}{dx} = 1.

    The integral becomes

    \int{u^{\frac{8}{3}}\,\frac{du}{dx}\,dx}

     = \int{u^{\frac{8}{3}}\,du}

     = \frac{3}{11}u^{\frac{11}{3}} + C

     = \frac{3}{11}(x - 2)^{\frac{11}{3}} + C.
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  4. #4
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    Quote Originally Posted by Defunkt View Post
    Let u = (x-2)^2 \Rightarrow dx = du.

    Then: \int (x-2)^\frac{8}{3} ~dx = \int u^{\frac{8}{3}} ~ du

    Can you solve it now?
    Attempt:
     \frac{3}{11} (x-2)^\frac{11}{3}+C

    Is this the answer?
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  5. #5
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    Quote Originally Posted by mj.alawami View Post
    Attempt:
     \frac{3}{11} (x-2)^\frac{11}{3}+C

    Is this the answer?
    Yes, Well done!
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