I have been trying to evaluate the integral: $$\int \frac{x}{\sqrt[12]{x^2+x+1}}\ dx$$ I approached the integral with trig substitutions, by completing the square and with integration by parts. I also tackled it as if it were a binomial integral, but nothing seems to work. In each case I get stuck and end up with an integral that I am not able to evaluate. Can someone give me a hint? Thank you.

  • $\begingroup$ Why do you believe this can be integrated in elementary terms? $\endgroup$
    – user414998
    Feb 12, 2017 at 7:35
  • $\begingroup$ Do you have a good reason to suspect that the integral can be expressed in terms of elementary functions? $\endgroup$ Feb 12, 2017 at 7:35
  • $\begingroup$ I found the integral in my calculus book.. $\endgroup$
    – user372003
    Feb 12, 2017 at 7:38
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    $\begingroup$ Whch calculus book is this? And how is it treating the hypergeometric functions that appear in this indefinite integral? $\endgroup$ Feb 12, 2017 at 7:44
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    $\begingroup$ @Denis This indefinite integral is expressed in terms of hypergeometric functions. If this is an introductory calculus book perhaps it is a typo as this cannot be solved by a substitution or integration by parts. $\endgroup$
    – user275377
    Feb 12, 2017 at 7:47

1 Answer 1


First we can extract the integrable part :


But then we have $I=\int\frac{dx}{\sqrt[12]{x^2+x+1}}$

From $x^2+x+1=(x+\frac12)^2+\frac34\quad$ by substituting $\quad\sinh(u)=\frac{2}{\sqrt 3}(x+\frac 12)$

You get $I=\int (\frac{\sqrt 3}{2}\cosh(u))^\frac56du$

which as other people have said, cannot be expressed in term of usual set of elementary functions.

You may reduce the exponent to a simple square root by substituting $v^6=\cosh(u)$ but the $dv$ makes reappear some $v^{12}$.

$\int \cosh(u)^\frac56du=6\int\frac{v^{10}}{\sqrt {v^{12}+1}}dv\quad$ this last one being clearly hypergeometric.

  • $\begingroup$ @Sophie, sure, though I wanted to show the intermediate $\sinh$ substitution because in case of a typo and $\sqrt[12]{}$ is just $\sqrt{}$ then this lead to actually solving the integral. But good remark. $\endgroup$
    – zwim
    Feb 12, 2017 at 9:07

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