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.
2026-05-06 04:10:44.1778040644
Trivial integration...
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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.