Solving the Integral ∫dx/(1-x)

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Homework Help Overview

The discussion revolves around solving the integral ∫dx/(1-x), with the original poster presenting two different methods that yield seemingly different results. The subject area is calculus, specifically focusing on integration techniques and properties of logarithmic functions.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster describes two methods for solving the integral, both leading to different expressions involving logarithms. Participants question the validity of the results and explore the implications of absolute values in the context of the solutions.

Discussion Status

The discussion includes attempts to clarify the apparent discrepancy in the results. Some participants suggest evaluating the expressions at specific values to understand the differences better. There is a focus on the properties of absolute values rather than a consensus on the methods themselves.

Contextual Notes

The original poster's question highlights a potential misunderstanding regarding the properties of logarithmic functions and absolute values, which is central to the discussion but remains unresolved.

Fernando Rios
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Homework Statement
Solve the following integral
Relevant Equations
∫dx/(1-x)
I solved the integral by two different methods and I get different answers.

Method 1:
∫dx/(1-x) = -∫-dx/(1-x), u=1-x, du=-dx

∫dx/(1-x) = -∫du/u = -ln|u| = -ln|1-x|

Method 2:
∫-dx/(x-1) = -∫dx/(x-1), u=x-1, du=dx

∫-dx/(x-1) = -∫du/u = -ln|u| = -ln|x-1|

What am I doing wrong?
 
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Try calculating both quantities for, say, ##x=3##.
 
Fernando Rios said:
Homework Statement:: Solve the following integral
Relevant Equations:: ∫dx/(1-x)

I solved the integral by two different methods and I get different answers.

Method 1:
∫dx/(1-x) = -∫-dx/(1-x), u=1-x, du=-dx

∫dx/(1-x) = -∫du/u = -ln|u| = -ln|1-x|

Method 2:
∫-dx/(x-1) = -∫dx/(x-1), u=x-1, du=dx

∫-dx/(x-1) = -∫du/u = -ln|u| = -ln|x-1|

What am I doing wrong?
At heart, your question really isn't about calculus -- it's about a property of the absolute value.
|a - b| = |b - a|, right?
 
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Fernando Rios said:
What am I doing wrong?
Nothing
 
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