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Math Help - An interesting indefinite integral

  1. #1
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    An interesting indefinite integral

    solve  \int \left[ \frac{ \arctan(x) }{ x - \arctan(x) } \right]^2 ~dx
    Last edited by mr fantastic; June 1st 2009 at 03:06 AM. Reason: Re-sized the square brackets
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    solve  \int \left[ \frac{ \arctan(x) }{ x - \arctan(x) } \right]^2 ~dx
    Maple gives the answer readily

    Spoiler:

     <br />
\frac{1 + x \tan^{-1}x}{\tan^{-1}x - x}<br />
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  3. #3
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    let \tan^{-1} x = t. then your integral becomes: I=\int \frac{t^2 \sec^2t}{(\tan t - t)^2} \ dt=\int \frac{t^2}{(\sin t - t\cos t)^2} \ dt=\int \frac{t}{\sin t} \cdot \frac{t \sin t}{(\sin t - t \cos t)^2} \ dt. now put u=\frac{t}{\sin t} and \frac{t \sin t}{(\sin t - t \cos t)^2} \ dt = dv.

    then du = \frac{\sin t - t \cos t}{\sin^ 2 t} \ dt, and v=\frac{-1}{\sin t - t\cos t}. thus integration by parts gives us: I=\frac{-t}{(\sin t - t \cos t) \sin t} + \int \frac{dt}{\sin^2 t}=\frac{-t}{(\sin t - t \cos t)\sin t } - \cot t + C. \ \ \ \ \ \ \ \ \ (1)

    to write the answer in terms of x, we have: \cot t = \frac{1}{x} and \frac{-t}{ (\sin t - t \cos t)\sin t}=\frac{-t(1 + \cot^2 t)}{1 -t \cot t}=\frac{-(1+x^2)\tan^{-1} x}{x(x - \tan^{-1} x)}. put these two in (1) and simplify to get the answer that

    danny and Maple already gave us.
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    Wonderful !
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