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Math Help - Integration Problem Has Me Stumped

  1. #1
    Newbie
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    Oct 2006
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    Integration Problem Has Me Stumped

    One question on my assignment has me stumped. I've tried a number of different methods, but every time, I either get stuck somewhere, or the equation quickly gets ridiculously out of hand. I'm not sure where to start here.

    \int \frac {x^2}{\sqrt[]{2x^2+4x+25}}dx

    I've used google and some math help sites to try to point myself in the right direction, but it hasn't been much help. I'm getting answers like this;





    Which both seem very wrong, and don't help at all.

    Can someone give me some advice?
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  2. #2
    Super Member
    Joined
    Jan 2009
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    Hi , so what is your answer , let's compare it with the answers you have got in the sites .

    Let me show that the two results are the same :

    We can see that both of them have this thing :

     \frac{1}{8} [ 2(x-3) \sqrt{2x^2 + 4x + 25 }]

    so let's compare  \log\left( \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23} } + \sqrt{ \frac{( \sqrt{2}x +\sqrt{2})^2}{23} + 1 }\right)  and  \sinh^{-1}\left( \sqrt{\frac{2}{23}} (x+1)  \right)


    In fact , we have  \sinh^{-1}(u) = \ln( u + \sqrt{u^2 + 1 } )

    by letting  u =  \sqrt{\frac{2}{23}} (x+1)  =  \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23}}

    We have  \log\left( \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23} } + \sqrt{ \frac{( \sqrt{2}x +\sqrt{2})^2}{23} + 1} \right) = \sinh^{-1}\left( \sqrt{\frac{2}{23}} (x+1)  \right)  so actually the two answers you've got are the same .
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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