How do you integrate 3ye^3z dz using u substitution?

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Homework Statement



Needing to integrate 3ye^3z dz

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The Attempt at a Solution


I believe you use u substitution and u=3z du=3dz
Then you get yze^3z. Is this correct?
 
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If you differentiate yze3z with respect to z do you get what you started with? I don't think so, but can you fix it?
 
>>Then you get yze^3z. Is this correct?
No completely, you have an extra factor of z
 
I believe differentiating ye^3z with respect to z will give 3ye^3z. Therefore, integrating 3ye^3z dz will be ye^3z. Correct?
 
Yes.
 
Sweet,thanks to the both of you
 
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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