Integration by parts and simplifying

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Homework Help Overview

The discussion revolves around the integration of the function \(\int \frac{\ln(x)}{x^2} dx\), specifically focusing on the method of integration by parts and the simplification of the integral. Participants are exploring their understanding of integration techniques in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply integration by parts but expresses confusion about the integration of \(1/x^2\). Some participants suggest viewing \(1/x^2\) as a polynomial to facilitate integration. Others confirm the integration result and discuss verification methods.

Discussion Status

The discussion is active, with participants providing guidance on the integration process. There is acknowledgment of the original poster's approach, and some clarification on integrating \(1/x^2\) has been offered. However, there is no explicit consensus on the final outcome of the integration.

Contextual Notes

Participants are navigating through the integration process while adhering to homework constraints, which may limit the sharing of complete solutions. The original poster expresses frustration, indicating the problem's complexity and their struggle to find clarity.

trajan22
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Hi,
I have been working on this problem for the longest time and have just run in circles with it. I am thinking the answer is obvious but for some reason I am missing it. I need to find [tex]\int \frac{ln(x)}{x^2} dx[/tex] I know that I need to use integration by parts and have tried a number of things, however the only way that the integral seems to be simplified is if I use this set up
u=ln(x) du=1/x
v=? dv=1/(x^2)
but from here I cannot integrate 1/x^2. Am i even on the right track with this one or is there an easier way? someone please help as this problem is truly annoying me.
 
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you're on the right track, but try thinking of 1/x^2 as polynomial, ie: x^(-2).
I'm sure you can integrate that.
 
oh yeah that's right...ok so if i integrate that then its simply -(1/x) correct?
 
trajan22 said:
oh yeah that's right...ok so if i integrate that then its simply -(1/x) correct?

thts correct

(if you are unsure try differentiating (-1/x))
 
oh yeah that's right...I always forget that its really easy to check these types of problems...thanks for all the help
 
I : lnx/x^2 dx
I: (-lnx/x) - I(1/x^2) dx
I: -lnx/x + 1/x + K
 

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