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Help me with this integral please:

  1. Aug 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find:

    2. Relevant equations

    [tex]\int \frac{dx}{1+x^{\frac{1}{4}}}[/tex]

    3. The attempt at a solution

    I tried partial fractions and substitution, did work.
    Tried to do it with a contour integral didn't work.
    please show all working.
     
  2. jcsd
  3. Aug 11, 2012 #2
    Do we split the integral by completing the square?
     
  4. Aug 11, 2012 #3

    micromass

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    Can you show us EXACTLY what you tried??

    No, we will not. Read the forum rules. We will NOT provide you with a solution. You will have to find it yourself using our hints.
     
  5. Aug 11, 2012 #4
    [tex]\int \frac{1}{(1+\sqrt[4]{x})^{2}-2\sqrt{x}}[/tex]
    then partial fractions to split the integral but it got tedious and im not sure if im on the right track, i need the solution asap, please help me.
     
  6. Aug 11, 2012 #5

    micromass

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    You can simplofy the integral to an integral of a rational functon

    [tex]\int \frac{1}{1+\sqrt[4]{x}}dx[/tex]

    by applying a substitution. Do you see an easy substitution that you can do?

    Please don't post stuff like this. If you want help fast, you should have posted sooner. It's not nice of you to push people like this.
     
  7. Aug 11, 2012 #6
    No, not really.
    What would you use as a substitution?
     
  8. Aug 11, 2012 #7

    micromass

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    Take [itex]u=\sqrt[4]{x}[/itex]. After that substitution, it should be easy.
     
  9. Aug 11, 2012 #8
    I think I was able to do it with a substitution.

    [tex]\int \frac{1}{1+x^{\frac{1}{4}}} dx \\ $Let u^4=x $ \therefore 4u^3 du = dx \\ I=4 \int \frac{u^3}{1+u} du\\ = 4\int u^2-u+1-\frac{1}{1+u} du $ by long division or synthetic division $ \\ = 4 \int [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du = 4[\frac{x^{\frac{3}{4}}}{3}-\frac{x^{\frac{1}{2}}}{2}+x^{\frac{1}{4}}-\ln ( x^{\frac{1}{4}}+1)]+C[/tex]
     
  10. Aug 11, 2012 #9
    [tex]\int \frac{1}{\sqrt{e^{2x}-1}} dx[/tex]

    Can be done in about five easy lines with a smart substitution.

    But I don't know what to use as a substitution, can you please help me?
     
  11. Aug 11, 2012 #10
    Do I let u=e^x?
     
  12. Aug 11, 2012 #11

    dextercioby

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    Well, just like above, get rid of the 'nasty' part, in this case, the exponential.
     
  13. Aug 11, 2012 #12

    micromass

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    That last line doesn't make sense. You can't say

    [tex]4\int u^2-u+1-\frac{1}{1+u} du= 4 \int [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du[/tex]

    You're not taking the integral anymore in the right hand side. You should write

    [tex]4\int u^2-u+1-\frac{1}{1+u} du= 4 [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)][/tex]

    But anyway, the solution is correct!
     
  14. Aug 11, 2012 #13

    micromass

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    Please start a new thread for a new problem.
     
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