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Math Help - indefinite integral

  1. #1
    Super Member Random Variable's Avatar
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    indefinite integral

    Challenge Problem:

     \int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx

    Moderator approved CB
    Last edited by CaptainBlack; December 4th 2010 at 10:36 AM.
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  2. #2
    Super Member Random Variable's Avatar
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    My solution is only valid for certain intervals.
    Last edited by Random Variable; December 4th 2010 at 03:26 PM.
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  3. #3
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    For the interval x>0, we can use the u-substitution
    u=\frac{x^2-1}{x^2+1}, so that
    x^2=\frac{1+u}{1-u},
    x=\frac{\sqrt{1+u}}{\sqrt{1-u}},
    dx=\frac{du}{\sqrt{1+u}(1-u)^{3/2}}=\frac{du}{(1-u)\sqrt{1-u^2}}.
    Thus,
    \int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx=\int\frac{u}{\sqrt{\le  ft(\frac{1+u}{1-u}\right)^2+1}}\frac{du}{(1-u)\sqrt{1-u^2}}
    \;\;=\int\frac{u(1-u)}{\sqrt{(1+u)^2+(1-u)^2}}\frac{du}{(1-u)\sqrt{1-u^2}}
    \;\;=\int\frac{u}{\sqrt{2+2u^2}\sqrt{1-u^2}}\,du
    \;\;=\frac{1}{\sqrt2}\int\frac{u}{\sqrt{1-u^4}}\,du
    \;\;=\frac{1}{2\sqrt2}\int\frac{2u}{\sqrt{1-u^4}}\,du
    \;\;=\frac{1}{2\sqrt2}\sin^{-1}u^2+C
    \;\;=\frac{1}{2\sqrt2}\sin^{-1}\left(\frac{x^2-1}{x^2+1}\right)^2+C

    --Kevin C.
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  4. #4
    Super Member Random Variable's Avatar
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    My solution looks different, but it only differs by a constant.

     \int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx

     = \int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx \ \frac{1/x}{1/x} \ \text{if} \ x \ne 0

     = \int \frac{(x - 1/x)}{x(x+ 1/x) \sqrt{x^{2}+1/x^{2}}} \ dx \ \text{if} \  x>0

     = \int \frac{(x - 1/x)}{x (x+ 1/x) \sqrt{(x+1/x)^{2} -2}} \ dx

     \text{let} \ x =e^{u}

    =\int \ \frac{(e^{u} - e^{-u})}{(e^{u}+ e^{-u}) \sqrt{(e^{u}+e^{-u})^{2} -2}} \ du

     \text{let}\ v=e^{u}+e^{-u}

      = \int \frac{1}{v \sqrt{v^{2} -2}} \ dv

     \text{let} \ v = \sqrt{2} \sec w

     = \frac{1}{\sqrt{2}} w + D

     = \frac{1}{\sqrt{2}} \sec^{-1} \Big(\frac{x + 1/x}{\sqrt{2}} \Big) + D

    where  D = C + \frac{\sqrt{2} \pi}{8}}
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  5. #5
    Super Member Random Variable's Avatar
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    But the function we're integrating is even, so for x<0,      \int \frac{x^{2}-1}{(x^{2}+1)\sqrt{x^4+1}} \ dx = - \frac{1}{\sqrt{2}} \sec^{-1} \Big(\frac{|x| + 1/|x|}{\sqrt{2}} \Big) + D , no?
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  6. #6
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    hadn't even seen this section about challenge problems, it's pretty interesting.

    perhaps this is one of the hardest integral i've ever seen, but with a little of magic we can take it down easily.

    note that

    \displaystyle\int{\frac{{{x}^{2}}-1}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}\,dx}=\int{\frac{1-\frac{1}{{{x}^{2}}}}{\left( x+\frac{1}{x} \right)\sqrt{{{\left( x+\frac{1}{x} \right)}^{2}}-2}}\,dx},

    whereat the substitution x+\dfrac1x=\dfrac{\sqrt2}t will solve the problem.
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