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In a set of additional excercises, I am asked to find $$\int \frac{1+x^2}{\arctan(x)}\,dx,$$ but I don't know how to solve it. I noticed it is of the form $\int \frac{1}{f(x)f'(x)}\,dx$ with $f(x)=\arctan(x)$, but this doesn't help me to solve it.

Is it possible that this is some sort of typo (even wolfram alpha was unable to solve this), with the idea that the intended excercise was $\int \frac{1}{(1+x^2)\arctan(x)}\,dx$ with solution $\ln|\arctan(x)|+C$?

Anton V.
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    Yes, it might be a typo. – Anik Bhowmick Oct 07 '18 at 15:00
  • @ParclyTaxel This is not a textbook problem, it comes from a new set of additional excercises from some guy who coaches students in maths, so he might not have checked them rigorously. Also, your tone is quite disrespectful. – Anton V. Oct 07 '18 at 15:11
  • @AntonV. But still, it qualifies as some exercise set. I presume there is an intended solution to the problem, so if there turns out to be no solution, it is probably a typo. – Parcly Taxel Oct 07 '18 at 15:12

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using the substitution $x=\tan(u)$ we can obtain: $$I=\int\frac{(1+\tan^2u)\sec^2u}{u}du=\int\frac{\sec^4u}{u}du$$ then you could try and expand this using series, but like you said it has no elementary antiderivative so the likelihood is that this was a mistake.

Henry Lee
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