Integrate x/sqrt(a^2+b^2-2abx)

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Homework Help Overview

The discussion revolves around the integration of the expression x/sqrt(a^2+b^2-2abx) with respect to x. Participants are exploring methods to approach this integral, particularly through substitution techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand a substitution method mentioned in a textbook. Some participants provide hints related to integration techniques, while others express uncertainty about the approach.

Discussion Status

Participants are actively engaging with the problem, sharing hints and discussing the appropriateness of methods. There is acknowledgment of the need for further exploration of the integration process, but no consensus has been reached on a specific solution.

Contextual Notes

There are indications that the original poster may have posted in an incorrect forum, which could affect the focus of the discussion. Additionally, the reliance on a textbook for guidance suggests that there may be assumptions about prior knowledge or methods that are not fully articulated in the thread.

albega
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Homework Statement


I need to integrate x/sqrt(a^2+b^2-2abx) with respect to x.

The Attempt at a Solution


This follows from a substitution of the form x=cost in a textbook I'm reading - they jump to the solution straight from the above and I have no idea how to go about it - any hints will be very gratefully appreciated, thanks :)
 
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albega said:

Homework Statement


I need to integrate x/sqrt(a^2+b^2-2abx) with respect to x.

The Attempt at a Solution


This follows from a substitution of the form x=cost in a textbook I'm reading - they jump to the solution straight from the above and I have no idea how to go about it - any hints will be very gratefully appreciated, thanks :)

Hint: [tex]\int \frac{Au + B}{\sqrt u}\,du = \int Au^{1/2} + Bu^{-1/2}\,du.[/tex]
 
pasmith said:
Hint: [tex]\int \frac{Au + B}{\sqrt u}\,du = \int Au^{1/2} + Bu^{-1/2}\,du.[/tex]

Got it, thanks! Was that something you could spot or did you just know it from experience - I don't think I would ever have thought of something like that...
 
You also should make an attempt to post HW in the correct HW forum. There is a perfectly good Calculus-HW forum listed right below this one, which is where this post belongs.
 
SteamKing said:
You also should make an attempt to post HW in the correct HW forum. There is a perfectly good Calculus-HW forum listed right below this one, which is where this post belongs.

Thread moved to Calculus HH. :smile:
 

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