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Solve,

$$ \int\frac{1}{1+x\ln^{2}x}dx $$

I have tried using $ \phi = e^{x} $ as a substituent... didn't quite work out for me. I don't need an answer, I just need hints. Yes, I've tried searching up online but to no avail.

Deltaspace
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lzrus
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    Going by Wolfram the integral has no expression in terms of elementary functions. Is there a reason you otherwise would expect it to have one, or is there a certain kind of answer you expect that allows for them? Or perhaps you could tell us where this integral comes from or why you need it? (For instance, definite integrals even for cases like these can have exact values using other techniques - the antiderivative isn't always the way to go!) – PrincessEev Dec 11 '20 at 02:17
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    That doesn't look trivial at all. In fact, I would go far as to say that it does not integrate into elementary functions. – Doug M Dec 11 '20 at 02:18

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What makes it seem "trivial" to you? In fact it is not trivial at all, for that it doesn't have elementary close forms.

Using substitution $t=\ln x$ and the original integral(denoted as $I$) becomes $$ I = \int \frac{1}{e^t t^2+1} \, d(e^t) = \int \frac{e^t}{e^t t^2+1} \, dt = \int \frac{1}{t^2+e^{-t}} \, dt $$

Substitute $p=-t$ and we get $$ I = -\int \frac{1}{p^2+e^p} \, dp $$ Which has no elementary closed forms according to answers in this question.

FFjet
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  • I'd always assumed questions i get stuck with tend to be "trivial" in the sense that it's rather easy to resolve... I'm aware in mathematics "trivial" has a different meaning. I apologise if I made any offence. Sorry. In any case, thank you for your time and kind assistance :) – lzrus Dec 11 '20 at 06:43