What Is the Correct Substitution for the Integral of 4x/(1+4x^2)?

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Homework Help Overview

The discussion revolves around the integral of the function 4x/(1+4x^2), with participants exploring appropriate substitution methods for solving the integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the integral and has tried a substitution with y=4x^2, which they found unhelpful. Other participants suggest using y=1+4x^2 as a potential substitution. There is also a mention of relating the integral to the derivative of a function.

Discussion Status

Participants are actively discussing different substitution strategies. Some have provided alternative approaches, and there is a sense of collaborative exploration without a clear consensus on the best method yet.

Contextual Notes

There are indications of confusion regarding the correct substitution and the relationship between the integral and logarithmic functions. The original poster's attempts and the responses suggest a need for clarification on the substitution process.

sparkle123
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How do you get that the integral of 4x/(1+4x^2) = 1/2*ln(1_4x^2) ?
I have no clue. I tried u substitution with y=4x^2 but that got me nowhere.
Thanks!
 
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Try substitution with y=1+4x^2.
 
Note that:

[tex] \frac{4x}{1+4x^{2}}=\frac{1}{2}\frac{8x}{1+4x^{2}}=\frac{1}{2}\frac{f'(x)}{f(x)}[/tex]

Where [tex]f(x)=1+4x^{2}[/tex]
 
Thanks hunt_mat and Dick! :)
 
sparkle123 said:
How do you get that the integral of 4x/(1+4x^2) = 1/2*ln(1_4x^2) ?
I have no clue. I tried u substitution with y=4x^2 but that got me nowhere.
Thanks!

Even though you could have included the 1 in the substitution, the one you chose would have solved your problem the same way since

[tex]\int \frac{du}{1+u} = \ln (1+u) + C[/tex] , whenever u>0.
 
Cool thanks!
 

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