Integrating Rational Functions: Solving ∫(x/(x-1)^3)

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Homework Help Overview

The discussion revolves around the integration of the rational function ∫(x/(x-1)^3). Participants are exploring various methods to approach this integral, particularly focusing on substitution techniques.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss attempts at u-substitution, with some expressing confusion over the effectiveness of different substitutions. There are also considerations of how to handle the numerator in relation to the substitution.

Discussion Status

There is an ongoing exchange of ideas, with some participants suggesting that a simpler perspective might yield a solution. Others are clarifying the notation used in their expressions, indicating a focus on ensuring accurate communication of mathematical ideas.

Contextual Notes

Some participants note the importance of proper notation and parentheses in mathematical expressions to avoid misinterpretation. There is also a sense of varying confidence in the complexity of the integral, with differing opinions on its difficulty.

whatlifeforme
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Homework Statement


integrate the following:

Homework Equations


∫(x/(x-1)^3

The Attempt at a Solution


i've tried u-substitution, finding an inverse trig function that matched the formula, and still can't figure out how to solve this problem.

u-subtitution for u=x gives the same problem. u-subsitution for x-1 gives du =1 which does not match the problem.
 
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whatlifeforme said:

Homework Statement


integrate the following:


Homework Equations


∫(x/(x-1)^3


The Attempt at a Solution


i've tried u-substitution, finding an inverse trig function that matched the formula, and still can't figure out how to solve this problem.

u-subtitution for u=x gives the same problem. u-subsitution for x-1 gives du =1 which does not match the problem.

u=(x-1) gives du=dx which does match the problem. If you are worried about the x in the numerator, if u=x-1, then x=u+1.
 
Its a fairly easy integral..dont go all complicated when you can't find the answer just stay on the ground bro :-p sometimes few problems can be easily solved if you just view it from a different angle

Dick said:
u=x-1, then x=u+1.

this should work. you would get u+1/u^3 du which you split and integrate
 
Hysteria X said:
Its a fairly easy integral..dont go all complicated when you can't find the answer just stay on the ground bro :-p sometimes few problems can be easily solved if you just view it from a different angle



this should work. you would get u+1/u^3 du which you split and integrate

No, you will not get [tex]u + \frac{1}{u^3},[/tex] which is what you wrote! If you really mean [tex]\frac{u+1}{u^3},[/tex] use parentheses, like this: (u+1)/u^3.
 
Hysteria X said:
this should work. you would get u+1/u^3 du which you split and integrate

Good advice but use more parentheses. You could easily mistake u+1/u^3 for u+(1/u^3) when you meant (u+1)/u^3.
 
Ray Vickson said:
No, you will not get [tex]u + \frac{1}{u^3},[/tex] which is what you wrote! If you really mean [tex]\frac{u+1}{u^3},[/tex] use parentheses, like this: (u+1)/u^3.

lol thanks for pointing it out..my bad :shy:
 

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