Intergration by Parts (IbP) problem

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Discussion Overview

The discussion revolves around solving the integral \(\int_{1}^{2e} x^2(\ln x)^{2} dx\) using integration by parts (IbP). Participants explore the steps involved in applying IbP and seek clarification on the correctness of their approach.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents their initial setup for IbP, defining \(u = (\ln x)^{2}\) and \(dv = x^{2} dx\), and calculates \(du\) and \(v\).
  • Another participant suggests integrating by parts again on the resulting integral from the first application.
  • A participant seeks confirmation on whether their initial steps are correct and if they should proceed with another application of IbP on the right-hand side.
  • Another participant affirms that the approach is correct and implies that integrating \(\ln(x)\) again should be straightforward if previously encountered.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the initial steps taken in the integration process and the suggestion to apply IbP again. However, the discussion does not resolve whether there are alternative methods or nuances in the integration process.

Contextual Notes

There are no explicit limitations or unresolved mathematical steps mentioned, but the discussion reflects a reliance on prior knowledge of integrating logarithmic functions.

Who May Find This Useful

Students or individuals interested in integration techniques, particularly those studying calculus or working on similar integral problems.

Ravenatic20
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[tex]\int_{1} ^{2e} x^2(ln x)^{2} dx[/tex]

I need to solve this using IbP. I made the following:

[tex]u = (ln x)^2[/tex]

[tex]du = (\frac{2 ln x}{x}) dx[/tex]



[tex]dv = x^2 dx[/tex]

[tex]v = \frac{x^3}{3}[/tex]

So I get:
[tex](ln x)^2 (\frac{x^3}{3}) \|_{1} ^{2e}[/tex][tex]- \int_{1} ^{2e} (\frac{x^3}{3}) 2 ln x dx[/tex]
(not sure how to make this look right)

Is this right?
Where do I go from here? Thanks
 
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Integrate by parts again.
 
So what I have is right so far? I just IbP on the RHS?
 
Yup, if you've integrated ln(x) before, it should be obvious that you can integrate it again.
 

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