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Different way to do this integral

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Make a substitution to express the integrand as a rational
    function and then evaluate the integral.
    [tex]\int \frac{x^{3}}{\sqrt[3]{x^{2}+1}}dx[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I didn't know what to substitute using their directions, but I got the answer (I think), could someone tell me if my answer is right and how to do it using the directions.
    [tex]u=x^{2} \\du=2xdx[/tex]
    [tex]\int \frac{u}{\sqrt[3]{u+1}} du[/tex]
    [tex]v=u+1 \\dv=du[/tex]
    [tex]\int \frac{v-1}{\sqrt[3]{v}} dv = \int v^{\frac{2}{3}}-v^{\frac{-1}{3}}dv[/tex]
    Therefore the answer is:
    [tex]\frac{3(x^{2}+1)^{\frac{5}{3}}}{5}-\frac{3(x^{2}+1)^{\frac{2}{3}}}{2}+C[/tex]
     
  2. jcsd
  3. Feb 10, 2013 #2

    Dick

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    I think you mislaid a 1/2 factor. But why not substitute u=x^2+1 to begin with?
     
  4. Feb 10, 2013 #3
    Yeah that works too, and I did forget the 1/2. Yeah I could do the u=x^2+1 too, idk what I was thinking lol. But is there any other way to do it using the directions?
     
  5. Feb 10, 2013 #4

    Dick

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    If you want to rationalize it completely try u^3=x^2+1.
     
  6. Feb 11, 2013 #5
    What does du become then?
    [tex]du=\frac{2}{3}\frac{x}{(x^2+1)^{\frac{2}{3}}}[/tex]?

    Idk what I'd do with that :\
     
  7. Feb 11, 2013 #6

    Dick

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    3u^2du=2xdx. The usual routine. Just keep going.
     
  8. Feb 11, 2013 #7
    Am I going about this the right way:
    [tex]u^{3}=x^{2}+1\\3u^{2}du=2xdx\\\frac{3}{2} \int \frac{(u^{3}-1)(u^{2})}{u}du[/tex]
    =[tex]\int u^{4}-udu[/tex]
     
  9. Feb 11, 2013 #8

    Dick

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    Looks ok to me. In terms of x it's the same answer you got before, right?
     
  10. Feb 11, 2013 #9
    Seems like a much easier way to do this. I didn't know you could take the derivative of each side of the u-substitution like that also.

    Thanks for the help :)
     
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