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How do I solve this?

$$\int \frac{\text{d}x}{x^2 + x \ln x}$$

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Have you tried changing the variable to $t=\ln x$? – Dennis Gulko May 3 '11 at 6:27
Wolfram: (no result found in terms of standard mathematical functions), so others don't have to check. – fdart17 May 3 '11 at 6:33
@Dinesh - see [here](… and here – Juan S May 3 '11 at 6:49
@Dinesh, yes, there is a theory of integration in closed form (that's what the software is based on). will get you started. @Doman; one solves a problem; one computes an antiderivative. – Gerry Myerson May 3 '11 at 6:52
Is there a purpose of computing the integral? You might consider a series expansion before doing the integration. – AD. May 3 '11 at 8:05

I haven't gotten the answer - but I've just noticed that there is something close. In particular:

$$\int \frac{(1 + x)\text{d}x}{x^2 + x \ln x} = \ln(x + \ln(x)) + \kappa$$

Which isn't it, but it then prompts the follow-up question - do we know how to perform the following?

$$ \int \frac{\text{d}x}{x + \ln x}$$


Firstly, this is way later, but I happen to come across a good response to this question. Whether or not it is 'elementary' is debateable, but here we are.

The aforementioned simplification that are to consider $$ \int \frac{\text{d}x}{x + \ln x}$$

This has $\text{li} (x) - \dfrac{x}{\ln x} + \dfrac{x}{\ln x + x}$ as an antiderivative. Of course, calling $\text{li} (x)$ elementary may be considered a little cheap, but I think it's a good result nonetheless. I attach this, from W|A, as a quick verification that this is a reasonable antiderivative.

I'm sorry for the necro! As an aside - what is the current rationale for posting second answers as opposed to editing first answers?

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check main post comments("Cplayer:no result found in terms of standard mathematical functions").But we can rewrite it to this form: $\int \frac{(1 + x)\text{d}x}{x^2 + x \ln x} = \int \frac{\text{d}x}{x^2 + x \ln x} + \int \frac{(1+x)(-2x-\ln x-1)\text{d}x}{(x^2 + x \ln x)^2}$ – Doman May 6 '11 at 8:52
@Doman: To be clear - Cplayer noted that W|A didn't find any result in its database. Fortunately, there are a vast amount of integrals that W|A cannot do that people nonetheless can do. This is not to claim that every integral has a meaningful antiderivative - I simply note that W|A is not the end-all and be-all of symbolic integration. – mixedmath May 6 '11 at 13:41

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