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Math Help - Definite Integral

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    Definite Integral

    \displaystyle \int_{0}^{\infty}\frac{\ln(x)}{x^2+2x+4}dx
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    Re: Definite Integral

    Have you tried integration by parts?

    Of course, romsek's solution works if all you need is the answer - you will have to put the correct polynomial in the denominator.

    - Hollywood
    Last edited by hollywood; April 27th 2014 at 11:14 PM.
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    Re: Definite Integral

    My Trial solution:: \int_{0}^{\infty}\frac{\ln(x)}{x^2+2x+4}dx = \int_{0}^{\infty}\frac{\ln(x)}{(x+1)^2+\left(\sqrt  {3}\right)^2}dx

    Now Let (x+1) = \sqrt{3}\tan \theta, Then dx = \sqrt{3}\sec^2 \theta d\theta

    So Integral convert into \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{\ln\left  (\sqrt{3}\tan \theta - 1\right)}{3\sec^2 \theta}\cdot \sqrt{3}\sec^2 \theta d\theta = \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}  {2}}\ln\left(\sqrt{3}\tan \theta - 1\right)d\theta

    \displaystyle = \frac{1}{\sqrt{3}}\cdot \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\ln\left(\frac  {\sqrt{3}\sin \theta - \cos \theta}{\cos \theta}\right)d\theta

    \displaystyle = \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}  {3}}\ln \left(\sqrt{3}\sin \theta -\cos \theta\right)d\theta - \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}  {3}}\ln (\cos \theta)d\theta

    Now How can I solve after that...

    Help me

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