Integration results in natural log of a natural log

  • Thread starter cmajor47
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  • #1
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SORRY FOR THE LACK OF FORMATTING, I SPENT 30 MINUTES TRYING TO FORMAT AND IT KEPT GETTING MESSED UP.


Homework Statement


Evaluate by substitution.
Integral of dx/xln(x^2)

Homework Equations


integral of 1/u du = ln(u) + C


The Attempt at a Solution


u = ln(x^2)
du = 2/x

Integral of dx/xln(x^2) = 1/2 ln(ln(x^2)) + C

The natural log of a natural log is confusing me. Did I not solve the problem correctly? If it is correct, is there a simpler way of forming the answer?
 

Answers and Replies

  • #2
I like Serena
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SORRY FOR THE LACK OF FORMATTING, I SPENT 30 MINUTES TRYING TO FORMAT AND IT KEPT GETTING MESSED UP.

Don worry about it.
Your formatting looks fine, better actually then some of the problem statements I've seen around.
And if you want, just ask a couple of questions around, and you'll see that ppl will help you improve your formatting.

The Attempt at a Solution


u = ln(x^2)
du = 2/x

Integral of dx/xln(x^2) = 1/2 ln(ln(x^2)) + C

The natural log of a natural log is confusing me. Did I not solve the problem correctly? If it is correct, is there a simpler way of forming the answer?

You are making the right substitution with u = ln(x2).
However, I think your substitution is slightly off.

You mention du = 2/x. However, you do not need du in your substitution, but dx.
So you should solve u = ln(x2) for x, and then take the derivative from that.

And yes, this is the simplest way to solve the problem.
 
Last edited:
  • #3
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Interestingly enough, it can be made to look a bit simpler. You could wrap the x^2 bit into the constant. This is realized in one of two ways. Start the integration by taking the square outside of the logarithm. You will arrive at 1/2 ln(ln(x)) + C. Or use log rules:

[tex]\frac{1}{2} ln(ln(x^2)) + C = \frac{1}{2} ln(2 ln(x)) + C = \frac{1}{2} ln(ln(x)) + \frac{1}{2} ln(2) + C = \frac{1}{2} ln(ln(x)) + C[/tex]

I know that wasn't the simplification you might have been looking for, but it is a bit easier.
 

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