Wolfram gave me one answer, examiners gave me another

  • Thread starter Saibot
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In summary, the conversation discusses the use of u substitution in integration and the resulting different expressions for the constant C. It is explained that both expressions are correct, as they differ only by a constant. The example of cos(2x) is given as an illustration of how different approaches to integration can yield different results.
  • #1
Saibot
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Homework Statement
Find the integral of 1/(2x-2)
Relevant Equations
u substitution
I removed the coefficient of 1/2 before integrating. So I had:
1/2 integral[(1/x-1)]
= 1/2 ln(x-1) + C

Using u substitution without removing the coefficient yields
1/2 ln(2x-2) + C

I get how it works both ways, and my answer must be wrong, but what exactly is causing this conflict? Am I not supposed to remove the factor of 1/2?

What is happening here?
 
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  • #2
##\ln(2x-2) = \ln(2(x-1)) = \ln(x-1) + \ln 2##
Therefore, ##\ln(x-1)## and ##\ln(2x-2)## differ only by a constant so both (divided by 2) are primitive functions of ##1/(2x-2)##.
 
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  • #3
What is happening is that as @Orodruin explained both are antiderivatives (=indefinite integrals) of the given function, however you use in both expressions the same letter ##C## for the constant and that is what is causing the confusion. We know that ##C## can be anything so it doesn't matter if you have just ##\ln x+C## in one expression and ##\ln x+C+\ln2## in the other, because if ##C## is anything, then so is ##C+\ln2## anything.
 
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  • #4
Got it, thanks guys. So I'm not actually wrong, we'll both just get different expressions for C.
 
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  • #5
Saibot said:
Got it, thanks guys. So I'm not actually wrong, we'll both just get different expressions for C.
It's happens quite often with integration. For example:
$$\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x -1 = 1 - 2\sin^2 x$$Any one of these may result depending on how you tackle the integral $$-2 \int \sin(2x)\ dx$$
 
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Related to Wolfram gave me one answer, examiners gave me another

1. Why did Wolfram give me one answer while examiners gave me another?

There are a few possible reasons for this discrepancy. One possibility is that Wolfram used a different set of data or assumptions in their calculations. Another possibility is that the examiners used a different methodology or set of criteria to evaluate your answer. It is also possible that there was an error in either the Wolfram answer or the examiner's answer.

2. Which answer should I trust, Wolfram's or the examiners'?

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4. Can I use Wolfram's answer as a reference in my research or assignments?

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5. How can I reconcile the differences between Wolfram's answer and the examiners' answer?

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