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How to integrate $$\int \frac{1}{\sqrt[4]{1+x^4}} dx $$

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    $\begingroup$ Ask Wolfram Alpha. Then decide if there are chances you will be able to get this answer yourself and if yes, what kind of things one should apply. $\endgroup$ – Start wearing purple Aug 23 '13 at 14:36
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$$ \begin{aligned} \int\dfrac{\mathrm{d}x}{\sqrt[\scriptstyle{4}]{x^4 + 1}} &=\int\frac{1}{x\,\sqrt[\scriptstyle{4}]{1 + 1/x^4}}\,\mathrm{d}x \end{aligned} $$ Now, set $ u=\sqrt[\scriptstyle{4}]{1+1/x^4} $. Then, $$ u^4 - 1 = \frac{1}{x^4} $$ $$ 4u^3\,\mathrm{d}u = -4\frac{1}{x^4}\frac{1}{x}\,\mathrm{d}x $$ $$ \frac{\mathrm{d}x}{x} = -\frac{u^3}{u^4 - 1}\,\mathrm{d}u $$ $$ \begin{aligned} \int\frac{\mathrm{d}x}{\sqrt[\scriptstyle{4}]{x^4+1}}&=-\int\frac{u^2}{u^4 - 1}\,\mathrm{d}u\\ &=-\int\frac{u^2 + 1 - 1}{\left(u^2 - 1\right)\!\!\left(u^2 + 1\right)}\,\mathrm{d}u\\ &=\int\frac{1}{1-u^2}\,\mathrm{d}u -\frac{1}{2} \int\frac{1}{u^2+1}-\frac{1}{u^2-1}\,\mathrm{d}u\\ &=\frac{1}{2}\int\frac{1}{1-u^2}\,\mathrm{d}u - \frac{1}{2}\arctan u + C\\ &=\frac{1}{4}\ln\left|\frac{u+1}{u-1}\right| - \frac{1}{2}\arctan u + C\\ &=\frac{1}{4}\ln\left|\frac{\sqrt[\scriptstyle{4}]{1+x^4}+x}{\sqrt[\scriptstyle{4}]{1+x^4}-x}\right| - \frac{1}{2}\arctan\!\left( \frac{\sqrt[\scriptstyle{4}]{1+x^4}}{x}\right) + C \end{aligned} $$

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  • $\begingroup$ Wolfram Alpha says it is right. link $\endgroup$ – user149844 Sep 28 '14 at 17:40
  • $\begingroup$ My bad, I am not a smart person. +1 $\endgroup$ – Frédéric Sep 28 '14 at 17:47
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Maple says $ x {\rm hypergeom}([1/4, 1/4], [5/4], -x^4)$. This can be converted to MeijerG or JacobiP or various Heun functions, but nothing elementary.

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