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Math Help - Another Integral w/ integration by parts

  1. #1
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    Question Another Integral w/ integration by parts

    Hello, here is the problem I am working on \int x^2arctan(2x)dx I set u=\ tan^{-1}{2x},\ du=\ \frac {2}{1+4x^2},\ dv=\ x^2,\ and\ v=\ \frac {x^3}{3}
    Now, I got \frac {x^3tan^{-1}{2x}}{3}-\int \frac {2x^3}{3+12x^2}dx How can I integrate \int \frac {2x^3}{3+12x^2}dx?
    Thanks,
    Matt
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Assuming that your work is okay, then just put z=3+12x^2.
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  3. #3
    Super Member Showcase_22's Avatar
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    I did this (minus the constants):




    In another post I found the "adding 0" method helpful. I tried it with this one and it gave me another answer. I'm not sure if they're the same but are written in different forms though:

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  4. #4
    Super Member 11rdc11's Avatar
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    Use long division and get

    \frac{x}{6} - \frac{x}{24x^2+6}

    You should be able to finish up from here
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