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Math Help - tricky integral (by parts)

  1. #1
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    tricky integral (by parts)

    integral of x(arctanx)^2

    I started doing it by parts but it goes on forever! any ideas on how to solve would be great
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  2. #2
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    Quote Originally Posted by khood View Post
    integral of x(arctanx)^2

    I started doing it by parts but it goes on forever! any ideas on how to solve would be great
    Let  x = \tan(z)

     z = \arctan(x)

     \frac{dz}{dx} = \frac{1}{x^2+1}

     dx = x^2+1dz = \tan^2(z)+1dz

     \int{z^2 \tan(z)(\tan^2(z)+1)\,dz}

     \int{z^2 \tan^3(z)+z^2\tan^2(z)\,dz}

     \int{z^2( \tan^3(z)+\tan^2(z))\,dz}

    There's still going to be a lot of integration by parts, but it's easier now, at least!
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    thanks
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  4. #4
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    Quote Originally Posted by khood View Post
    integral of x(arctanx)^2

    I started doing it by parts but it goes on forever! any ideas on how to solve would be great
    You're right, you can use integration by parts.

    Let u = (\arctan{x})^2 \implies du = \frac{2\arctan{x}}{x^2 + 1}

    and dv = x \implies v = \frac{x^2}{2}.

    Remembering that \int{u\,dv} = uv - \int{v\,du} we get


    \int{x(\arctan{x})^2\,dx} = \frac{(x\arctan{x})^2}{2} - \int{\frac{x^2\arctan{x}}{x^2 + 1}\,dx}.

    To evaluate this integral, you'd need to long divide, to get

    \frac{x^2\arctan{x}}{x^2 + 1} = \arctan{x} - \frac{\arctan{x}}{x^2 + 1} to give

     \frac{(x\arctan{x})^2}{2} - \int{\frac{x^2\arctan{x}}{x^2 + 1}\,dx}

     =  \frac{(x\arctan{x})^2}{2}- \int{\arctan{x}\,dx} - \int{(\arctan{x})\frac{1}{x^2 + 1}\,dx}.

    Then use the substitution z = \arctan{x} to get

    \frac{(x\arctan{x})^2}{2}- x\arctan{x}
     - \frac{1}{2}\log{(1 + x^2)} - \int{z \,dz}

    Can you go from here?
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