Solving Confusing Integral Equations

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Homework Help Overview

The discussion revolves around solving a complex integral equation involving the variable x and its limits. The integral in question is expressed as int[x/((L/2)+d-x)^2], which has led to confusion among participants regarding its setup and evaluation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express frustration over the integral's complexity and the presence of x in both the upper limit and as a dummy variable. There are attempts to clarify the integral's form and to propose substitutions to simplify the expression.

Discussion Status

Several participants are actively questioning the validity of the integral's setup and exploring different substitution methods. While some suggestions for simplification have been made, there is no clear consensus on the best approach or solution at this point.

Contextual Notes

There is an ongoing discussion about the implications of having the variable x in both the upper limit and as a dummy variable, which raises questions about the integral's formulation. Participants are also navigating through various substitution methods without arriving at a definitive resolution.

acedeno
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Homework Statement



int[x/((L/2)+d-x)^2] * the integral of x over [(L over 2 + d - x) all squared]*


Homework Equations


Integral chart



The Attempt at a Solution


I have done this so many ways with so many different answers. Could somebody who is without a doubt sure of the answer please respond because this integral is driving me INSANE!
 
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You can't have x in the upper limit and as an integration dummy variable.
 
Not following.
 
Your dummy variable is x, your upper limit contains x.
 
acedeno said:
Not following.
Is this the integral?

[itex]\displaystyle \int\frac{x}{((L/2)+d-x)^2}\ dx[/itex]
 
SammyS said:
Is this the integral?

[itex]\displaystyle \int\frac{x}{((L/2)+d-x)^2}\ dx[/itex]

Yes, that is the integral.
 
To simplify things, Let A = (L/2) + d .

Your integral becomes: [itex]\displaystyle \int\frac{x}{(A-x)^2}\ dx[/itex]

Notice the (A - x)2 = (x - A)2.

Use the substitution: u = x - A .
 
No, use the substitution:

[tex] u = (x - A)^{2}[/tex]
 
Dickfore said:
No, use the substitution:

[tex] u = (x - A)^{2}[/tex]

No unique substitution, but Sammy's substation is easier to work with
 

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