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I need to solve the indefinite integral: $$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx,$$ but I can not find any technique which could solve it.

Can you help me please?

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  • $\begingroup$ Is this the correct form of the integrand function ? The $-4+2$ under the roots seems a bit odd. $\endgroup$ – Rebellos Jan 17 '18 at 18:29
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Hint: $$(x^2+x^{-2})^2=x^4+x^{-4}+2.$$

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It is :

$$(x^2+x^{-2})^2=x^4+x^{-4}+2$$

which means that your initial integral can be substituted to :

$$\int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} dx = \int\frac{\sqrt{(x^2+x^{-2})^2}}{x^3}dx = \int \frac{x^2 + x^{-2}}{x^3}dx$$

since $x^2 + x^{-2} \geq 0 \space \forall x \in \mathbb R$.

So :

$$\int \frac{x^2 + x^{-2}}{x^3}dx= \int \bigg(\frac{1}{x} + \frac{1}{x^5}\bigg)dx=\int x^{-1}dx + \int x^{-5}dx= \ln|x| +-4x^{-4}+ C$$

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