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Math Help - Integration #1

  1. #1
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    Integration #1

    f(x) = xtan^{-1}x

    Find f'(x) and hence find \int_{0}^{1} tan^{-1}(x) dx.

    I can find f'(x), but I don't know how to hence find the integral. We've covered (in the context of this forum) basic integration techniques in class, but not inverse tan. I looked up how to do it in general, but my confusion arises from the wording -- hence -- which suggests that I should be able to figure it out from the derivative. Any help is greatly appreciated.
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  2. #2
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    Re: Integration #1

    Quote Originally Posted by Fratricide View Post
    f(x) = xtan^{-1}x

    Find f'(x) and hence find \int_{0}^{1} tan^{-1}(x) dx.

    I can find f'(x), but I don't know how to hence find the integral. We've covered (in the context of this forum) basic integration techniques in class, but not inverse tan. I looked up how to do it in general, but my confusion arises from the wording -- hence -- which suggests that I should be able to figure it out from the derivative. Any help is greatly appreciated.
    do this by parts

    if you know it that is

    \int udv= uv-\int vdu



    u=tan^{-1}(x); du = \frac{1}{x^2+1}dx, dv=dx, v=x

    that's how I'd solve the integral
    Last edited by Jonroberts74; July 24th 2014 at 11:15 PM.
    Thanks from topsquark and Fratricide
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  3. #3
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    Re: Integration #1

    Let's see:
    f(x) = x*arctan(x) so
    f'(x) = arctan(x) + x/(1+x^2)
    Then,
    arctan(x) = f'(x) - x/(1+x^2)
    Integrating both sides with respect to x, we should have
    S arctan(x)dx = f(x) - (1/2)ln(1+x^2)
    Now, plug in x=0 and 1 and do the subtraction. xD
    Thanks from Fratricide
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