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Thread: An interesting indefinite integral

  1. #1
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    An interesting indefinite integral

    solve $\displaystyle \int \left[ \frac{ \arctan(x) }{ x - \arctan(x) } \right]^2 ~dx $
    Last edited by mr fantastic; Jun 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 $\displaystyle \int \left[ \frac{ \arctan(x) }{ x - \arctan(x) } \right]^2 ~dx $
    Maple gives the answer readily

    Spoiler:

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

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

    to write the answer in terms of $\displaystyle x,$ we have: $\displaystyle \cot t = \frac{1}{x}$ and $\displaystyle \frac{-t}{ (\sin t - t \cos t)\sin t}=\frac{-t(1 + \cot^2 t)}{1 -t \cot t}=\frac{-(1+x^2)\tan^{-1} x}{x(x - \tan^{-1} x)}.$ 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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