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How can we evaluate this difficult integral:

$$\int\frac{\tan^2x}{1-x\tan^2x}dx=\int\frac{1}{\cot^2x-x}dx$$

I tried using integration by parts but I suppose it doesn't work with this.

In the worst case, if this integral doesn't have a closed formula, how can we $prove$ that it cannot be found?

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  • $\begingroup$ I doubt it would have a closed form $\endgroup$ – Simply Beautiful Art Dec 3 '17 at 16:01
  • $\begingroup$ You can't. $\endgroup$ – Paracosmiste Dec 3 '17 at 16:02
  • $\begingroup$ Do you really need to find an explicit antiderivative of $\frac{1}{\cot^2 x-x}$, or the original question is about the evaluation of a definite integral? $\endgroup$ – Jack D'Aurizio Dec 3 '17 at 18:39
  • $\begingroup$ My question was just to find the explicit antiderivate. $\endgroup$ – user507623 Dec 3 '17 at 20:24
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It is interesting to note that your integral telescopes, starting with

$$\int\frac{\tan^2x}{1-x\tan^2x}dx=\int\frac{\tan^2x}{1-x\tan^2x}\frac{(1+x\tan^2x)}{(1+x\tan^2x)}dx$$

Next multiply top and bottom by $1+x^2\tan^4x$ and so on. In this process the denominator tends to 1 (assuming the integral is between the limits of $x=0$ and $x=\pi/4$) and the numerator to the infinite series $S$

$$S=\tan^2 x+x\tan^4 x+x^2\tan^6x+x^3\tan^8x+x^4\tan^{10}x+...$$

Now anybody who claims there is a nice closed form to be found can be given the task of integrating these terms one by one until all faith in this proposition runs dry.

I understand that this is not a proof, but all you asked for was a way of demonstrating the difficulty with this integral.

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