How to evaluate $\int \frac{x}{\sqrt[12]{x^2+x+1}}\ dx$

Question

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.

Answer 1

First we can extract the integrable part :

$\int\frac{x\,dx}{\sqrt[12]{x^2+x+1}}=\frac12\int\frac{(2x+1)\,dx}{\sqrt[12]{x^2+x+1}}-\frac12\int\frac{dx}{\sqrt[12]{x^2+x+1}}=\frac{6}{11}(x^2+x+1)^\frac{11}{12}-\frac12\int\frac{dx}{\sqrt[12]{x^2+x+1}}$

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.