Integrating, probably by parts

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    Integrating parts
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Homework Help Overview

The discussion revolves around the integration of the expression \(\int{x(\ln{x})^3dx}\), with participants exploring different methods for solving it, particularly integration by parts and substitution techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts a substitution method involving \(u=\ln(x)\) and questions the validity of their approach after obtaining an unexpected result. Others suggest using integration by parts with specific choices for \(u\) and \(dv\), and some confirm that both methods can lead to valid results.

Discussion Status

The discussion is active, with participants providing different methods and confirming the effectiveness of the approaches discussed. There is acknowledgment that both the substitution and integration by parts methods are viable, though no consensus on a single preferred method has been reached.

Contextual Notes

Participants are navigating the complexities of integration techniques and the potential for confusion in applying substitutions correctly. The original poster expresses uncertainty about their method, indicating a need for clarification on integration rules.

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I have the expression [tex]\int{x(\ln{x})^3dx}[/tex]
I thought I had a quick way to integrate by parts but it turned out that I had accidentally evaluated [tex]\int{x\ln{x}dx}[/tex] instead.
Revisiting [tex]\int{x(\ln{x})^3dx}[/tex], I wanted to start by making a strange substitution, wherein u=ln(x), du=1/x dx, and x=e^u. This meant that when I rewrote the integral, instead of multiplying dx by a constant to get it to be du, I multiplied it by x (which in this case was e^u). Is that allowed? Because I got a very different, much uglier answer than the book's.

I'd appreciate any comments, whether on my weird "method" or on a more standard approach to evaluating [tex]\int{x(\ln{x})^3dx}[/tex]
 
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Try integration by parts with u = (ln(x))^3 and dv = x dx
 
Last edited:
Your substitution method should work fine. Your should be integrating [tex]\int{Exp[2u] u^3du}[/tex]. If you do it by integration by parts, you will need to do it 3 times.
 
wurth_skidder_23 said:
Try integration by parts with u = (ln(x))^3 and dv = x dx
thanks, that got me the book's answer!
 
And yes, the other way does work also. Nifty!
 

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