Partial Fraction Decomposition With Quadratic Term

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Discussion Overview

The discussion centers around the concept of partial fraction decomposition, specifically addressing the inclusion of both linear and quadratic terms in the decomposition of a rational function. Participants explore the reasoning behind the structure of the decomposition as presented in a calculus textbook example.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Homework-related

Main Points Raised

  • One participant questions why both ##x-1## and ##(x-1)^2## appear in the decomposition when the original denominator is ##(x-1)^2(x-2)##.
  • Another participant suggests that omitting the ##x-1## term would lead to a more complex numerator for the quadratic term, implying that including it simplifies the integration process.
  • A third participant expresses confusion and requests a more explicit explanation or additional resources for understanding the topic.
  • A later reply refers to a previous thread where this question was addressed, suggesting that participants review that discussion before continuing.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the explanation for the inclusion of the linear term in the decomposition, and some express confusion about the topic. The discussion remains unresolved with varying levels of understanding among participants.

Contextual Notes

There may be limitations in the explanations provided, as some participants seek further clarification and resources. The discussion does not fully address the underlying assumptions or mathematical steps involved in the decomposition process.

Cosmophile
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Hey, all! I'm learning partial fraction decomposition from Serge Lang's "A First Course in Calculus." In it, he gives the following example:

\int\frac{x+1}{(x-1)^2(x-2)}dx

He then decomposes this into the following sum:

\frac{x+1}{(x-1)^2(x-2)} = \frac{c_1}{x-1}+\frac{c_2}{(x-1)^2}+\frac{c_3}{x-2}

My question is this: On the right hand side (RHS), ##x-1## and ##(x-1)^2## appear. Why is this the case, when the original denominator only had the ##(x-1)^2##? I hope this makes sense, and any help here is greatly appreciated!
 
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If you left out the x-1 denominator term, then the numerator for (x-1)^2 would be a+bx. The expression you are given is equivalent and is easier to integrate.
 
I'm afraid I don't really understand. Could you explain more explicitly, or direct me to a good resource on this?
 
I've answer this in a previous thread, so read that first and then you can ask more questions here.
 

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