Integration Problem: Solving Compact Results in Electrodynamics

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SUMMARY

The integration problem presented involves the expression \(\int \frac{r dr}{(r^2 + a^2 - 2rau)^{3/2}}\), which results in the compact form \(\frac{ru - a}{a(1-u^2)\sqrt{r^2 + a^2 - 2rau}}\). This integration frequently arises in electrodynamics contexts, particularly in relation to problems found in David Griffiths' textbook. The discussion highlights the challenge of deriving this compact result from the integral, as the solution manual provides only the final answer without detailed steps.

PREREQUISITES
  • Understanding of integral calculus, specifically integration techniques.
  • Familiarity with electrodynamics concepts as presented in David Griffiths' textbook.
  • Knowledge of mathematical notation and manipulation of algebraic expressions.
  • Experience with solving physics problems that involve integration.
NEXT STEPS
  • Study integration techniques in depth, focusing on advanced methods such as substitution and integration by parts.
  • Review relevant sections in David Griffiths' "Introduction to Electrodynamics" to understand the context of the integration problem.
  • Explore Sadikus' textbook for alternative approaches to similar integration problems in electrodynamics.
  • Practice solving integrals that yield compact results to improve efficiency in deriving solutions.
USEFUL FOR

This discussion is beneficial for physics students, educators, and researchers specializing in electrodynamics, as well as anyone seeking to enhance their integration skills in the context of physical applications.

Psi-String
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Hi! Could someone give me an idea how the following integration be carried out??

\int \frac{r dr}{ \left( r^2 + a^2 - 2rau \right)^{3/2}} = \frac{ru - a}{a(1-u^2)\sqrt{r^2 + a^2 - 2rau}}

where u and a are constant.

I have encounter such integration several times in electrodynamics.

I can solve it in lengthy way, but can't carry out such compact result as above. Could someone help me?? Thanks a lot!
 
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Look up Sadikus textbook. Or David Griffiths. You'll find a good result there.
 
well, this is an integration I need to carry out when solving Griffiths' problem. The solution manual only give the result as above, it didn't take out step by step, and I don't think it is that obvious that we can solve it just by one equation XDDDD
 

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