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    wonderful integral

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    Quote Originally Posted by dapore View Post
    To start...

    \int{\left(\frac{e^x}{e^{3x} + 1}\right)^2\,dx} = \int{\frac{e^x}{(e^{3x} + 1)^2}\,e^x\,dx}.

    Now make the substitution u = e^x so that du = e^x\,dx and the integral becomes

    \int{\frac{u}{(u^3 + 1)^2}\,du}

    = \int{\frac{u}{[(u + 1)(u^2 - u + 1)]^2}\,du}

     = \int{\frac{u}{(u + 1)^2(u^2 - u + 1)^2}\,du}

    which can be solved using partial fractions.
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    Quote Originally Posted by dapore View Post
    Sub .   e^{3x} + 1 = \frac{1}{t}

     3e^{3x} ~dx = \frac{dt}{t^2}

    We have

     I = \frac{1}{3} \int_0^1 \left( \frac{ \sqrt[3]{t} }{  \sqrt[3]{1-t}}\right) ~dt

     = \frac{1}{3} \frac{ \Gamma(\frac{4}{3}) \Gamma(\frac{2}{3})}{\Gamma(2)}

     = \frac{1}{9} \Gamma(\frac{1}{3})\Gamma(\frac{2}{3})

     = \frac{1}{9} \frac{\pi}{\sin(\frac{\pi}{3}) } = \frac{2\pi}{9\sqrt{3}}


    ps : or straightforward computation for  I = \int_0^{\infty} \frac{u}{(u^3+1)^2} ~du

    Use  u \mapsto \frac{1}{u}

     I = \int_0^{\infty} \frac{u^3}{(u^3+1)^2}~du

    Integration by parts ,

     I = -\frac{1}{3} \left[ \frac{u}{u^3+1} \right]_0^{\infty}  + \frac{1}{3} \int_0^{\infty} \frac{du}{u^3+1}

     = \frac{1}{3} \int_0^{\infty} \frac{du}{u^3+1}

    Consider  J = \int_0^{\infty} \frac{du}{u^3+1}

    Use again u \mapsto \frac{1}{u} and we have

     J = \int_0^{\infty} \frac{u}{u^3+1}~du so

     J + J =  \int_0^{\infty} \frac{u+1 }{u^3+1}~du

     =  \int_0^{\infty} \frac{du}{u^2 - u + 1 }  = \frac{4\pi}{3\sqrt{3}}

     J = \frac{2\pi}{3\sqrt{3 }}

    Finally ,  I = \frac{1}{3} J = \frac{2\pi}{9\sqrt{3 }}
    Last edited by simplependulum; July 2nd 2010 at 10:41 PM.
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