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Math Help - Hard Integration Q

  1. #1
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    Hard Integration Q

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

    I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?
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  2. #2
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    You could try saying x^4 = tan^2 y

    dx = (sec^2 y)/[4(tan y)^{3/2}]

    x^2 = tan y

    Which would give you 1/4 times the integral of (tan y - 1)/[(tan y)^(3/2)(tan y + 1)]

    I am not sure if that will lead any where and you may need another substitution.
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  3. #3
    Junior Member eddie2042's Avatar
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    not so sure about the final answer but if you set

    <br /> <br />
u = x^2<br /> <br />

    you'll get

    <br /> <br />
\frac {1} {\sqrt{1 + u^2}}<br /> <br />

    and i know the antiderivative of that is

    <br /> <br />
\arcsin u<br /> <br />

    Good luck
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  4. #4
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    \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},

    substitute x+\frac1x=\frac{\sqrt2}t and your integral equals -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r  ight)+k.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},

    substitute x+\frac1x=\frac{\sqrt2}t and your integral equals -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r  ight)+k.
    You show 'em son.
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  6. #6
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    Wow, just wow. You are unbelievable!

    Quote Originally Posted by Krizalid View Post
    \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},

    substitute x+\frac1x=\frac{\sqrt2}t and your integral equals -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r  ight)+k.
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  7. #7
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    Quote Originally Posted by vuze88 View Post
    Find \int \left ( \frac{x^2-1}{x^2+1}.\frac{1}{\sqrt{1+x^4}} \right )dx

    I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?

    Or if you never think of the wonderful substitution  t = x - 1/x , you can choose this substitution  x = \tan(y) which is familiar with all of us .


     dx = \sec^2(y) dy

    the integral becomes :

     I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy

     = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~  dy

     = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~  dy

     = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy

     = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy

     = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy

    Haha , then substitute  \sin(2y) = t ~\implies 2\cos(2y)dy = dt

     I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}

     = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}})  + C

     t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1}

    so

     I  = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1})  + C
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  8. #8
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    cool, thanks for that simplependulum
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
    Or if you never think of the wonderful substitution  t = x - 1/x , you can choose this substitution  x = \tan(y) which is familiar with all of us .


     dx = \sec^2(y) dy

    the integral becomes :

     I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy

     = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~  dy

     = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~  dy

     = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy

     = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy

     = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy

    Haha , then substitute  \sin(2y) = t ~\implies 2\cos(2y)dy = dt

     I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}

     = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}})  + C

     t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1}

    so

     I  = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1})  + C
    You're a trooper for writing that out.
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