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Math Help - Integration Question

  1. #1
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    Integration Question

    In this question I do not see how x^5 becomes an x^4 in the second step.
    I understand that x^2 = u-4 and that x^2 * x^2 = x^4, but if we started with x^5 and we now have x^4, were is the extra x?


    x^5/ third root of x^2+4
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  2. #2
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    Quote Originally Posted by gammaman View Post
    In this question I do not see how x^5 becomes an x^4 in the second step.
    I understand that x^2 = u-4 and that x^2 * x^2 = x^4, but if we started with x^5 and we now have x^4, were is the extra x?

    x^5/ third root of x^2+4
    the "extra" x is part of "du" ...

    \int \frac{x^5}{\sqrt[3]{x^2+4}} \, dx

    u = x^2 + 4

    du = 2x \, dx

    x^2 = u - 4

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

    \frac{1}{2} \int \frac{(u-4)^2}{\sqrt[3]{u}} \, du

    \frac{1}{2} \int \frac{u^2 - 8x + 16}{\sqrt[3]{u}} \, du

    \frac{1}{2} \int u^\frac{5}{3} - 8x^{\frac{2}{3}} + 16x^{-\frac{1}{3}} \, du

    integrate and back substitute ...
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