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Simple integration based on chain rule

  1. Oct 10, 2006 #1
    I think I've got this, but I'm not quite sure, especially about multiplying to get du to be what I want. Can someone please tell me if this is correct? (The first line is the problem.) Thanks!

    [tex]\int\frac{3xdx}{\sqrt[3]{2x^2+3}}[/tex]
    I set [tex]u=2x^2+3, du=4x[/tex] so the problem becomes
    [tex]\int u^{-\frac{1}{3}}(\frac{3}{4})(\frac{4}{3})3xdx[/tex]
    rearranging the constants gives the much nicer equation
    [tex]\frac{3}{4}\int u^{-\frac{1}{3}}du[/tex]
    [tex]=(\frac{3}{4})(\frac{u^\frac{2}{3}}{\frac{2}{3}})+C[/tex]
    [tex]=\frac{9(2x^2+3)^\frac{2}{3}}{8}+C[/tex]
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 10, 2006 #2

    radou

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    Homework Helper

    Correct. Just one detail, though: du = 4xdx, but you probably just let it out by mistake, since the solution is correct.
     
  4. Oct 10, 2006 #3

    Office_Shredder

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    The best way to check your work is to just take the derivative, and see if it matches what you integrated in the first place
     
  5. Oct 10, 2006 #4
    Thank you very much!
    I can now happily continue my four sections of calc homework that I wisely saved for long lazy days...
     
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