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    wonderful integral

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

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

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

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

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

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

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

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

    We have

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

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

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

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


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

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

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

    Integration by parts ,

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

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

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

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

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

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

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

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

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