Integration results in natural log of a natural log

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?

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Homework Helper
SORRY FOR THE LACK OF FORMATTING, I SPENT 30 MINUTES TRYING TO FORMAT AND IT KEPT GETTING MESSED UP.

Your formatting looks fine, better actually then some of the problem statements I've seen around.

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.

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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:

$$\frac{1}{2} ln(ln(x^2)) + C = \frac{1}{2} ln(2 ln(x)) + C = \frac{1}{2} ln(ln(x)) + \frac{1}{2} ln(2) + C = \frac{1}{2} ln(ln(x)) + C$$

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