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Math Help - IntegrL of fraction

  1. #1
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    IntegrL of fraction

    Need to integrate z/1+4z^2 +z^4

    Please help. thanks!
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    your fractions don't make any sense.
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    Quote Originally Posted by Niall101 View Post
    Need to integrate z/1+4z^2 +z^4

    Please help. thanks!
    I assume ...

    \int \frac{z}{1+4z^2+z^4} \, dz


    \frac{z}{z^4+4z^2+4 - 3} = \frac{z}{(z^2+2)^2 - 3}<br />

    \int \frac{z}{(z^2+2)^2 - 3} \, dz

    u = z^2+2

    du = 2z \, dz

    \frac{1}{2} \int \frac{2z}{(z^2+2)^2 - 3} \, dz<br />

    \frac{1}{2} \int \frac{1}{u^2-3} \, du

    partial fractions ...

    \frac{1}{u^2-3} = \frac{1}{2\sqrt{3}} \left(\frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}}\right)


    \frac{1}{4\sqrt{3}} \int \frac{1}{u-\sqrt{3}} - \frac{1}{u+\sqrt{3}} \, du

    \frac{1}{4\sqrt{3}} \left[\ln(u-\sqrt{3}) - \ln(u+\sqrt{3})\right] + C

    \frac{1}{4\sqrt{3}} \ln\left(\frac{u-\sqrt{3}}{u+\sqrt{3}}\right) + C

    \frac{1}{4\sqrt{3}} \ln\left(\frac{z^2+2-\sqrt{3}}{z^2+2+\sqrt{3}}\right) + C
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    Thanks so much. Makes it look easy. I was trying to integrate by parts. Could I do something similar then for integral of

    1/(3Z^3 + 7)
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    Member Miss's Avatar
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    Quote Originally Posted by Niall101 View Post
    Thanks so much. Makes it look easy. I was trying to integrate by parts. Could I do something similar then for integral of

    1/(3Z^3 + 7)
    \int \frac{dz}{3z^3+7} is dangerous.
    See its result:
    integrate 1/(3z^3+7) - Wolfram|Alpha

    But you can do it as an infinite series.
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    Quote Originally Posted by Miss View Post
    \int \frac{dz}{3z^3+7} is dangerous.
    See its result:
    integrate 1/(3z^3+7) - Wolfram|Alpha

    But you can do it as an infinite series.
    Oh thats nasty. I dont think I should know how to do anything like that. Im trying to solve dy/dx = (7X - 3Y)^3

    And i used the substitution Z = 7X - 3Y

    and i come up with that integral



    Edit: I ended up with A SIMPLE ODE and i was treating it as variable seperable I think this was the problem
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    Member Miss's Avatar
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    Quote Originally Posted by Niall101 View Post
    Oh thats nasty. I dont think I should know how to do anything like that. Im trying to solve dy/dx = (7X - 3Y)^3

    And i used the substitution Z = 7X - 3Y

    and i come up with that integral
    Is this a problem in Differential Equations ?
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    Quote Originally Posted by Miss View Post
    Is this a problem in Differential Equations ?
    Yes i think I have solved now tho. Thanks!
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    Quote Originally Posted by Niall101 View Post
    Yes i think I have solved now tho. Thanks!
    Hmmm. Just wondering, how did you solve it?
    Did you change it to a separable equation by using the suggested substitution?
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    I tried to. Actually I just have a big mess now
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  11. #11
    Member Miss's Avatar
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    Quote Originally Posted by Niall101 View Post
    I tried to. Actually I just have a big mess now
    The problem is that you will face \int \frac{1}{7-3z^3} \, dz.
    Did try anything to solve it?
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    I tried substitution and integration by parts but I cant find the right substitutions to use.
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  13. #13
    Member Miss's Avatar
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    Quote Originally Posted by Niall101 View Post
    I tried substitution and integration by parts but I cant find the right substitutions to use.
    Am sure, its not easy. Its a dangerous integral.
    Are your sure that the exponent is 3 in the original question?
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    Quote Originally Posted by Miss View Post
    Am sure, its not easy. Its a dangerous integral.
    Are your sure that the exponent is 3 in the original question?
    Original Question is "By using a suitable substitution solve the differential equation"

    dy/dx = (7X - 3Y)^3
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