Why Does Integrating (x+2)/(x^2+4x) Lead to a Logarithmic Function?

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Homework Help Overview

The discussion revolves around the integration of the function (x+2)/(x^2+4x), with a focus on understanding why the result involves a logarithmic function. The subject area includes calculus, specifically integration techniques and properties of logarithms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to work through the integration using substitution and expresses confusion regarding the logarithmic result. Some participants suggest considering the appropriate substitution and explore the relationship between the integral of 1/u and the natural logarithm.

Discussion Status

The discussion is active, with participants providing guidance on substitution methods and clarifying the connection between the integral and logarithmic functions. There is an ongoing exploration of the reasoning behind the logarithmic outcome, but no consensus has been reached on the complete understanding of the integration process.

Contextual Notes

The original poster mentions a lack of confidence in their understanding of logarithms, which may influence their approach to the problem. There is also a reference to prior difficulties with logarithmic concepts from algebra.

vipertongn
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S (x+2)/(x^2+4x) dx

I've been learning about natural logs and their properties but at this answer I get befuddled at how to work it out. I think perhaps substitution, and some properties with logs but I am very weak in logs...Didn't do well in it when I was in algebra.
I want to know how to do this equation, although I know it has this as an answer:

log(abs(x^2+4x))/2

Thanks for any help.
 
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vipertongn said:
I think perhaps substitution

You think rightly. Now take a stab at choosing the right substitution. There aren't that many choices available, so this shouldn't be too hard.
 
hmmm ok let's see u = x^2+4x --> du=2x+4 --> du/2=x+2

makes the equation

1/2 S du/U but 1/u means ln|u| though right? why is it a log?
 
vipertongn said:
hmmm ok let's see u = x^2+4x --> du=2x+4 --> du/2=x+2

makes the equation

1/2 S du/U but 1/u means ln|u| though right? why is it a log?

That's the right substitution, but you're missing something in du.
u = x^2 + 4x ==> du = (2x + 4)dx

[tex]\int du/u = ln |u| + C[/tex]
An antiderivative of 1/u is ln |u | because the derivative of ln |u| is 1/u. It's a sort of inverse relationship, similar to the relationship between the equations y = ln(x) and x = e^y.
 

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