Solving the Integral ∫dx/(1-x)

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SUMMARY

The integral ∫dx/(1-x) can be solved using two methods, yielding results of -ln|1-x| and -ln|x-1|. Both methods are valid and represent the same mathematical expression due to the properties of absolute values. The discrepancy arises from the interpretation of the absolute value, as |1-x| is equivalent to |x-1|. Therefore, the solutions are consistent, and no error exists in the calculations.

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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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