How to Integrate 1/(1+2x+x^2) dx Without Getting Stumped

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    Dx Integrate
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Discussion Overview

The discussion revolves around the integration of the function 1/(1+2x+x^2) with respect to x. Participants are seeking assistance with the integration process, exploring different approaches and substitutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in integrating the function during an exam.
  • Another participant suggests that the integral can be rewritten as \(\int \frac{dx}{(1+x)^2}\), although the reasoning behind this transformation is not fully explained.
  • A later post proposes a substitution \(u = x + 1\), leading to a simpler integral \(\int u^{-2} du\), which may clarify the application of integration rules.
  • One participant presents an expression \(1 + x^2 + \frac{1}{3}x^3\), but it is unclear how this relates to the integration problem.

Areas of Agreement / Disagreement

There is no clear consensus on the integration method, as participants propose different approaches and transformations without resolving which is the most effective.

Contextual Notes

Participants have not fully explored the implications of their substitutions or transformations, and there may be missing steps in the integration process that could affect the outcome.

FUNKER
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In an exam i stumbled when i saw this q
integrate:
1/(1+2x+x^2) dx

help
 
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[tex]\int \frac{dx}{1+2x+x^2}[/tex]

[tex]\int \frac{dx}{(1+x)^2}[/tex]

Whats the pro
 
himanshu121 said:
[tex]\int \frac{dx}{1+2x+x^2}[/tex]

[tex]\int \frac{dx}{(1+x)^2}[/tex]

Whats the pro
?
?
 
[tex]1 + x^2[/tex] + [tex]\frac{1}{3}x^3[/tex]
 
Last edited:
Substitute [itex]u=x+1[/itex]. [itex]du=dx[/itex] so we get

[tex]\int\frac{dx}{(1+x)^2}=\int\frac{du}{u^2}=\int u^{-2}du[/tex]

That should make it a little easier to see what rule you can apply.
 

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