So, what is the problem asking for? Integration by Parts for ∫(z^3e^z)dz

narutoish
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Homework Statement


∫((z^3)(e^z ))dz

Homework Equations



I just tried u dv - ∫v du

The Attempt at a Solution



u = z^3 dv = e^z
du = 3z^2 v = e^z

= z^3e^z - ∫(3e^z (z^2)) dz

I got this far but after that if I try integration by parts again, it gets too confusing.
 
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You just have to keep going. Parts, parts and parts again!
 
narutoish said:

Homework Statement


∫((z^3)(e^z ))dz


Homework Equations



I just tried u dv - ∫v du


The Attempt at a Solution



u = z^3 dv = e^z
du = 3z^2 v = e^z

= z^3e^z - ∫(3e^z (z^2)) dz

I got this far but after that if I try integration by parts again, it gets too confusing.

Integrate ∫(3e^{z} (z^{2})) dz with integration by parts till you get the term∫e^{z}dz
 
narutoish said:

Homework Statement


∫((z^3)(e^z ))dz


Homework Equations



I just tried u dv - ∫v du


The Attempt at a Solution



u = z^3 dv = e^z
du = 3z^2 v = e^z

= z^3e^z - ∫(3e^z (z^2)) dz

I got this far but after that if I try integration by parts again, it gets too confusing.

You can avoid confusion by using some extra symbols. Let ##I = \int z^3 e^z \, dz##. Integration by parts gives you ##I = z^3 e^z - 3I_1##, where ##I_1 = \int z^2 e^z \, dz##. Now look at ##I_1## in the same way, etc. Using separate symbols like that helps to keep things straight and to reduce errors.
 
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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