How Do You Integrate x/(sqrt[1+x^2])?

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Discussion Overview

The discussion revolves around the integration of the function x/(sqrt[1+x^2]) dx. Participants explore various methods of integration, including substitution techniques and the relationship to known integrals.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states the integral they need to solve is x/(sqrt[1+x^2]) dx and expresses uncertainty about how to arrive at the answer, which they believe is sqrt[1+x^2].
  • Another participant comments on the substitution u = 1+x^2, noting that it transforms the integral into a form involving u^{-1/2}, which is a known integral.
  • A participant reflects on their confusion regarding the differential, confirming that if u = 1 + x^2, then du = 2x dx, and questions whether this leads to the integral being 0.5u^(-1/2).
  • A later reply affirms the correctness of the previous participant's reasoning and suggests integrating through the transformed expression.

Areas of Agreement / Disagreement

Participants generally agree on the substitution method and the transformation of the integral, but there is no consensus on the final steps or the clarity of the integration process.

Contextual Notes

Some participants express uncertainty about the integration steps following the substitution, indicating potential gaps in understanding the process.

Who May Find This Useful

Students or individuals seeking assistance with integration techniques, particularly those involving substitution methods in calculus.

Brewer
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I have to integrate:
x/(sqrt[1+x^2]) dx

I know the answer is sqrt[1+x^2], but I can't work out how to get there. I tried the substitution u = 1+x^2, but that didn't seem to get me any where. It also looks like the differential of arsinx, but its not quite. How do I get to the answer?

Any help would be greatly appreciated.

Thank you in advance
 
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Why didn't that subs help: it transforms it up to a scalar multiple to the integral of u^{-1/2}, which you know how to do.
 
i managed to confuse myself, when getting du.

if u = 1 + x^2

then du/dx = 2x

so du = 2x dx yes?

so does that mean that the integral is now 0.5u^(-1/2)?

I can't remember what to do when I get to that stage.

Thanks though
 
Brewer said:
i managed to confuse myself, when getting du.

if u = 1 + x^2

then du/dx = 2x

so du = 2x dx yes?

so does that mean that the integral is now 0.5u^(-1/2)?

I can't remember what to do when I get to that stage.

Thanks though

From what I can see, that is correct. Now simply integrate through:

[tex]\int\frac{du}{2\sqrt{u}}[/tex]

and follow through with your answer.
 
Ta. It makes more sense now.

Thankyou, both of you
 

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