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Thread: Simplifying an Integral

  1. #1
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    Simplifying an Integral

    How does this
    $\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}}$

    become
    $\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}}$

    According to my notes I let

    u = x^2 + 4
    x^2 = u - 4

    but I do not see how that gives me the above.
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  2. #2
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    Quote Originally Posted by gammaman View Post
    How does this
    $\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}}$

    become
    $\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}}$

    According to my notes I let

    u = x^2 + 4
    x^2 = u - 4

    but I do not see how that gives me the above.
    $\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}} \, dx$

    $\displaystyle u = x^2 + 4$

    $\displaystyle du = 2x \, dx$

    $\displaystyle x^2 = u - 4$

    $\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}} \cdot 2x \, dx$

    substitute ...

    $\displaystyle \frac{1}{2}\int\frac{(u-4)^2}{\sqrt[3]{u}} \, du$
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  3. #3
    Member Nacho's Avatar
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    put $\displaystyle u^3=x^2+4$ an get: $\displaystyle
    \frac{3}
    {2}\int {\frac{{\left( {u^3 - 4} \right)^2 }}
    {u} \cdot u^2 du}
    $
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  4. #4
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    what happens to the X^5 though? Do we just get rid of it? Why?
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  5. #5
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    Quote Originally Posted by gammaman View Post
    what happens to the X^5 though? Do we just get rid of it? Why?
    it's still there ... what is $\displaystyle \frac{1}{2} \cdot x^4 \cdot 2x$ ?
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