Why is ∫x/(1+x^2)dx = 1/2ln(1+x^2)?

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Discussion Overview

The discussion centers around the integral ∫x/(1+x^2)dx and seeks to understand why it equals 1/2ln(1+x^2). Participants explore integration techniques, particularly substitution methods, to clarify the process of solving this integral.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant asks for relevant integration rules to understand the integral.
  • Another participant suggests the substitution method as a potential approach.
  • A participant proposes using the substitution u = (1+x^2) and asks about the differential du.
  • Further clarification is provided on the substitution, detailing how it transforms the integral and leads to the natural logarithm function.
  • There is a reiteration of the substitution method as a fundamental technique that should be familiar to those studying integration.

Areas of Agreement / Disagreement

Participants generally agree on the utility of the substitution method, but there is no consensus on the specific integration rules or techniques that should be applied in this case.

Contextual Notes

Some participants assume familiarity with integration techniques, which may not be the case for all readers. The discussion does not resolve whether the substitution method is the only or best approach for this integral.

5P@N
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why is ∫x/(1+x^2) dx = 1/2ln(1+x^2)?

If you could just show me a couple of relevant integration rules, that would be great. I'm having trouble figuring this one out.
 
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Hi 5P:

Let u = (1+x^2). Then, what is du? With this substitution, what form does the integral then take?

Hope this helps.

Regards,
Buzz
 
It's the same substitution as before: ##u(x) = 1 + x^2## and therefore ## u' = du / dx = 2x ## which gives you ##x dx = \frac{1}{2} du## and then ##1/u## to integrate on ##u##, i.e. ##ln|u|##.
 
5P@N said:
If you could just show me a couple of relevant integration rules
You should already have seen the substitution method, which is one of the first techniques that are presented. Also, it's also one that you should try first when you have an integration problem. This technique might not be useful in some cases, but it's reasonably simple, so if it doesn't work, you haven't wasted much time.
 

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