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Fairly simple but incorrect integral

  • #1

Homework Statement



[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]

Homework Equations





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]
 

Answers and Replies

  • #2
33,486
5,174
Your equation for dx is wrong. If u3 = x -2, then 3u2du = dx.
 

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