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Math Help - Integrating Arctan!

  1. #1
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    Integrating Arctan!

    I can't seem to find a substitution for the inverse of tan anywhere, and online it looks like a mess. Can someone tell me the step by step process for integrating the following, please.

    x^2arctan(6x)dx

    Is this with integration by parts?

    Thank you!

    ..

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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Integrating Arctan!

    Yes, I would try integration by parts, where:

    u=\tan^{-1}(6x)\,\therefore\,du=\frac{1}{(6x)^2+1}\,dx

    dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3

    Then use long division on the integrand of the integral that results.
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    Re: Integrating Arctan!

    Quote Originally Posted by MarkFL2 View Post
    Yes, I would try integration by parts, where:

    u=\tan^{-1}(6x)\,\therefore\,du=\frac{1}{(6x)^2+1}\,dx

    dv=x^2\,dx\,\therefore\,v=\frac{1}{3}x^3

    Then use long division on the integrand of the integral that results.
    Thanks for your reply, just a question by the way, does it matter which term I choose as u or v when I'm integrating by parts?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Integrating Arctan!

    Yes, I use the LIATE rule to decide which will be u.
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    Re: Integrating Arctan!

    Quote Originally Posted by MarkFL2 View Post
    Yes, I use the LIATE rule to decide which will be u.
    After doing so, I'm left with:

    x^3arctan(6x)/3 - ∫x^3/(108x^2 + 3)

    (sorry for the messy notation)

    How do I go about integrating the second term?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Integrating Arctan!

    I would leave the second term as:

    -\frac{1}{3}\int\frac{x^3}{36x^2+1}\,dx

    You should find with polynomial division:

    \frac{x^3}{36x^2+1}=\frac{1}{36}\left(x-\frac{x}{36x^2+1} \right)
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Integrating Arctan!

    Sorry to interrupt, but I think that:

    d(\arctan(6x)) = \frac {6 dx}{(6x)^2 + 1}
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Integrating Arctan!

    Quote Originally Posted by ILikeSerena View Post
    Sorry to interrupt, but I think that:

    d(\arctan(6x)) = \frac {6 dx}{(6x)^2 + 1}
    Thanks, you are right, I got careless with the chain rule!
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