How Do You Solve ∫ln(x)dx?

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    Brain Integral
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Discussion Overview

The discussion revolves around the integration of the natural logarithm function, specifically the integral ∫ln(x)dx. Participants explore different methods for solving this integral, including integration by parts, and clarify some common confusions related to logarithmic integrals.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses confusion regarding the integral ∫ln(u)du and seeks assistance.
  • Another participant suggests using integration by parts, noting the derivative of x ln x.
  • A participant questions if the integral could simply be 1/u, indicating uncertainty about the notation and the integration process.
  • One participant corrects the misunderstanding about the integral of 1/u, clarifying that it results in ln(x) + c, not the other way around.
  • A later reply provides a detailed integration by parts approach, showing the steps to arrive at the integral of ln(x) using u = ln(x) and dv = dx.

Areas of Agreement / Disagreement

The discussion includes multiple viewpoints and some confusion regarding the integration process. There is no consensus on the initial misunderstanding, but a method for solving the integral is presented.

Contextual Notes

Participants exhibit varying levels of understanding regarding integration techniques, and some assumptions about notation and integration rules are not fully clarified.

KSCphysics
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One of my recent problems is with double integrals... and I am having a brain fart on the [tex]\int ln(u) du[/tex] can you do this?
 
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You can do it by parts. Consider that
[tex]\frac{d}{dx} x \ln x = \ln x + 1[/tex]
 
hrmm... i see what your saying..
 
wouldn't it just be 1/u?
unless that little dash is a negative sign mean your are integrating 1/ln(u) then I'm not sure but i think it would just be u then, but i am probably wrong
 
iluvsr20s,

You are probably confused with the fact that [tex]\int \frac{1}{u} du = \ln x + c[/tex]. It is not the other way around.
 
Pfft...

... hehe. this ain't too bad. Can we change it to int(ln(x),dx) though? It's how I've always done it notation-wise.

let u = ln(x), dv = dx --> du = 1/x(dx), v = x

So, u*v -int(v,du) = int(u,dv)

x*ln(x) - int(1,dx) = int(ln(x),dx)
x*ln(x) - x = int(ln(x),dx)

Et voila.
 

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