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Just an integral

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{dx}{1+\sqrt[3]{x-2}}[/tex]

    2. Relevant equations

    The answer is given: [tex]\frac{3}{2}(x-2)^\frac{3}{2}-3(x-2)^\frac{1}{3}+ln|1+(x-2)^\frac{1}{3}|+C[/tex]

    I have to get my answer to look just like this.

    3. The attempt at a solution

    [tex]u=x-2[/tex], [tex]du=dx[/tex]

    [tex]=\int\frac{dx}{1+\sqrt[3]{u}}[/tex]

    [tex]w=1+\sqrt[3]{u}[/tex]

    [tex]dw=\frac{du}{3u^\frac{2}{3}}[/tex]

    [tex]3u^\frac{2}{3}dw=du[/tex]

    [tex](w-1)^2=u^\frac{2}{3}[/tex]

    [tex]3(w-1)^2dw=du[/tex]

    [tex]=3\int\frac{(w-1)^2dw}{w}[/tex]

    [tex]=3\int\frac{w^2-2w+1}{w}dw[/tex]

    [tex]=3\int\frac{w^2}{w}dw-3\int\frac{2w}{w}dw+3\int\frac{1}{w}dw[/tex]

    [tex]=3\int\(wdw-6\int\(dw+3\int\frac{dw}{w}[/tex]

    [tex]=\frac{3}{2}(w^2)-6w+3ln|w|+C[/tex]

    [tex]=\frac{3}{2}(1+\sqrt[3]{u})^2-6(1+\sqrt[3]{u})+3ln|1+\sqrt[3]{u}|+C[/tex]

    [tex]=\frac{3}{2}(1+\sqrt[3]{x-2})^2-6(1+\sqrt[3]{x-2})+3ln|1+\sqrt[3]{x-2}|+C[/tex]

    From here I don't know where to go to get the answer stated above or if I'm not on the right track.
     
  2. jcsd
  3. Feb 15, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are doing fine. Now, just multiply the first two terms out. You can throw the constants out, since you already have a '+C'. Now combine what's left.
     
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