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Thread: need help with integration

  1. #1
    Feb 2009

    need help with integration

    Int (3x^3.5)/(3+x^2.5) dx

    I tried using integration by parts and the equation uv-int (vdu) and im not getting the right answer. Could someone please show me the steps on how to do this question?
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  2. #2
    Senior Member
    Dec 2007
    Anchorage, AK
    Have you considered making the u-substitution $\displaystyle u=\sqrt{x}$? Then $\displaystyle dx=2u\,du$, and your integral becomes
    $\displaystyle \int\frac{3x^{7/2}}{3+x^{5/2}}\,dx=\int\frac{6u^8}{3+u^5}\,du$
    $\displaystyle =\int{6u^3}-\frac{18u^3}{3+u^5}\,du$

    The first term is easily integrated, leaving the rational function $\displaystyle \frac{18u^3}{3+u^5}$ to be integrated.
    Using the fifth roots of unity, one can factor the denominator above into real linear and quadratic factors:
    $\displaystyle (u^5+3)=(u+3^{1/5})(u^2-3^{1/5}\frac{1-\sqrt5}{2}u+3^{2/5})(u^2-3^{1/5}\frac{1+\sqrt5}{2}u+3^{2/5})$

    One can then use partial fractions to separate the above into rational functions with linear or quadratic denominators, which you should be able to integrate.

    --Kevin C.
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