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Homework Help: Fairly simple but incorrect integral

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

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

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

    2. Relevant equations



    3. The attempt at a solution

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

    [tex]u=\sqrt[3]{x-2}[/tex]

    [tex]u^3=x-2[/tex]

    [tex]3udu=dx[/tex]

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

    [tex]w=1+u[/tex]

    [tex]w-1=u[/tex]

    [tex]dw=du[/tex]

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

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

    [tex]=3\int dw-3\int\frac{dw}{w}[/tex]

    [tex]=3w-3ln|w|+C[/tex]

    [tex]=3(1+u)-3ln|1+u|+C[/tex]

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

    [tex]=3+3(x-2)^\frac{1}{3}-3ln|1+(x-2)^\frac{1}{3}|+C[/tex]
     
  2. jcsd
  3. Mar 1, 2010 #2

    Mark44

    Staff: Mentor

    Your equation for dx is wrong. If u3 = x -2, then 3u2du = dx.
     
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