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Calculate the following integral $$\int_x^{2x} \frac{dt}{\sqrt{t^4+t^2+2}}$$

My approach: If complete the square, we have $$t^4+t^2+2=\left(t^2+\frac{1}{2}\right)^2-\frac{1}{4}+2=\left(t^2+\frac{1}{2}\right)^2+\frac{7}{4}$$

So, let $\left(t^{2}+\dfrac{1}{2}\right)=\sqrt{\dfrac{7}{4}}\tan(\theta)$, then $2\,dt=\sqrt{\dfrac{7}{4}}\sec^2(\theta) \, d\theta$. Replacing in the integral, we have $$\int \frac{dt}{\sqrt{t^4+t^2+2}} =\int \frac{\sec(\theta) \, d\theta}{2t}$$ How continues this??, I don't think this is the better way. Any help, plis.

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    $\begingroup$ @asdasdasd There is no elementary antiderivative for this integral. Where you perhaps asked to take the derivative of this integral instead (with respect to x)> $\endgroup$ – GaussTheBauss Dec 26 '15 at 1:21
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    $\begingroup$ @asdasdasd: Is the integral written correctly, because it is ugly or was it asking for a derivative? $\endgroup$ – Moo Dec 26 '15 at 1:22
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    $\begingroup$ There is a typo: The differential satisfies $2 t\,dt = \sqrt{\frac{7}{4}} \sec^2 \theta \,d\theta$. $\endgroup$ – Travis Dec 26 '15 at 1:26
  • $\begingroup$ @GaussTheBauss Sorry for the question, but How you know this integral doesn't have elementary antiderivative? $\endgroup$ – asdasd asd Dec 26 '15 at 1:32
  • $\begingroup$ @asdasdasd Well, you try for a couple minutes to find one, and then you realize it's impossible. So you go on the trusty Wolfram Alpha, and check. $\endgroup$ – GaussTheBauss Dec 26 '15 at 1:49
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The integrand does not possess an elementary anti-derivative. See Liouville's theorem and the Risch algorithm for more information. More to the point, non-trivial integrals of algebraic functions containing elements of the form $\sqrt{P(x)},$ where P is a cubic or quartic polynomial, are expressible in terms of elliptic integrals, if at all.

As an aside, we have $\displaystyle\int_0^\infty\frac{dx}{\sqrt{x^4+x^2+\color{red}1}}=K\bigg(\dfrac12\bigg),$ see OEIS A$249282$.

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