How to perform an integral with a polynomial and a radical?

ainster31
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Homework Statement



kuwnk.png


Homework Equations





The Attempt at a Solution



How did they go from the first step in the blue to the second step in the blue?

I tried integration by parts but that didn't work.
 
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Dick said:
Just u substitution. Put u=x^(2/3).

I suck at integration... that was such an easy one. =[

I tried for several papers to do substitution ##u=x^{1/3}##.
 
ainster31 said:

Homework Statement



kuwnk.png


Homework Equations





The Attempt at a Solution



How did they go from the first step in the blue to the second step in the blue?

I tried integration by parts but that didn't work.

They took a factor of x^(-2/3) outside of the radical where it becomes x^(-1/3). Then just u substitution u=x^(2/3). Sorry, I accidentally deleted my first post while editing. So this is out of order.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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