Integrating Rational Functions with Substitution: How to Solve Tricky Integrals

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Homework Help Overview

The discussion revolves around finding the integral of a rational function, specifically from 1/3 to 1/2 of the expression 1/x(x+1)(x-1)(x^2+1). Participants explore various substitution methods and the implications of partial fraction decomposition in the context of integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of substitution, with one suggesting u=x^4-1. There is uncertainty about the appropriate substitution and whether additional terms are needed in the numerator. Questions arise regarding the integration of the resulting expressions and the handling of limits.

Discussion Status

Several participants are actively engaging with the problem, offering suggestions and clarifications. There is a back-and-forth regarding the integration process, with some participants questioning assumptions and others providing insights into the decomposition of fractions. The discussion reflects a collaborative effort to navigate the complexities of the integral.

Contextual Notes

Participants note the importance of correctly applying limits after changing variables and the potential confusion that arises from improper notation. There is an emphasis on ensuring clarity in mathematical expressions throughout the discussion.

  • #31
Shannabel said:
nevermind, i wrote it down opposite from you :)

okay so now i have
1/4 * [ln(u)-ln(u+1)] between 1/3 and 1/2, right?

Yep!

EDIT: err, no. The limits are not right. :redface:
 
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  • #32
I like Serena said:
Yep!

EDIT: err, no. The limits are not right. :redface:

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81
= 1/4 [ln((-15/16)/(1/16))-ln((-80/81)/(1/81))]
= 1/4 [ln(-15)-ln(-80)]
=1/4 [ln(15/80)]
= 1/4ln(3/16)

:)
 
  • #33
Shannabel said:
what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81

The limits 1/3 and 1/2 would've been wrong if you used them while the integral was in terms of u, but you changed them right here and those are the correct limits; you didn't tell us you changed them in the previous post :wink:
 
  • #34
What Bohrok said! :approve:
 

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