Finding the indefinite integral

  • Thread starter csc2iffy
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  • #1
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Homework Statement


[itex]\int[/itex](x^(1/3)/(x^(1/3)+1))dx


Homework Equations


I know I have to use u substitution
u=x^(1/3)
du=1/(3x^(2/3))dx


The Attempt at a Solution


I know that the denominator of the equation will be u+1, but I don't understand how to find the numerator because I thought it would just be u, but wolfram alpha says it's u^3?
 

Answers and Replies

  • #2
Dick
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Homework Statement


[itex]\int[/itex](x^(1/3)/(x^(1/3)+1))dx


Homework Equations


I know I have to use u substitution
u=x^(1/3)
du=1/(3x^(2/3))dx


The Attempt at a Solution


I know that the denominator of the equation will be u+1, but I don't understand how to find the numerator because I thought it would just be u, but wolfram alpha says it's u^3?

If x^(1/3)=u then x^(2/3)=u^2. So du=(1/(3*u^2))dx. Does that tell you where the extra two powers of u are coming from?
 

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