How do I solve this?
$\int \frac{\text{d}x}{x^2 + x \ln x}$
How do I solve this?
$\int \frac{\text{d}x}{x^2 + x \ln x}$
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}$
------------UPDATE-------------
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?