Integrate (5x+2)dx/(x-2) from 0->1

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Discussion Overview

The discussion revolves around the integration of the function (5x+2)/(x-2) from 0 to 1. Participants explore various methods for solving the integral, including substitution and integration by parts, while expressing challenges in finding a suitable approach.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to split the integral into two parts: (5x)/(x-2) and (2)/(x-2) but struggles to find a solution.
  • Another participant suggests using the substitution u=x-2 and questions which substitutions have been tried.
  • A different participant proposes integration by parts with u=(5x+2) and dv=(1/(x-2)), seeking clarification on how this method would work.
  • One participant claims that the suggested u substitution simplifies the integral significantly, allowing it to be integrated using the power rule.
  • Another participant expresses confusion about the process and reflects on their understanding by stating "feck I am dumb."

Areas of Agreement / Disagreement

Participants express varying opinions on the effectiveness of different substitution methods, and there is no consensus on the best approach to solve the integral.

Contextual Notes

Some participants have not fully detailed their attempts at substitution, and there may be unresolved steps in the integration process.

erjkism
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i am trying to integrate Integrate (5x+2)dx/(x-2) from 0 to 1. i have tried splitting the equation up into (5x)/(x-2) + (2)/(x-2)
i have tried every U substitution i can think of but i can't figure out how to do it.
 
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Which substitutions have you tried? What about u=x-2?

Note in future that you should post these sort of questions in the suitable homework forum, and no the technical forums.
 
sorry about that
 
integration by parts

u=(5x+2) dv=(1/(x-2))

cristo said:
Which substitutions have you tried? What about u=x-2?

Note in future that you should post these sort of questions in the suitable homework forum, and no the technical forums.

how would that work?
 
Last edited:
The u substitution suggested above completely kills the problem ice. The integral becomes a linear polynomial in u over u, and then can be broken into pieces and integrated by the power rule.
 
ice109 said:
how would that work?

Let u=x-2, then du=dx and x=u+2. This converts the integral into \int\frac{5(u+2)du}{u}=\int\left[5+\frac{10}{u}\right]du
 
cristo said:
Let u=x-2, then du=dx and x=u+2. This converts the integral into \int\frac{5(u+2)du}{u}=\int\left[5+\frac{10}{u}\right]du

feck I am dumb
 

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