$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attempts, i could not solve further.
I think this function is not integrable.Am i correct?If not,how should i move ahead.Please help.

Edit: with some reasoning, and without controversial part
I see no way of calculating this primitive using human tools like integration by parts, substitution and so on. However, inspired by the example here, we could try a function in the form
$$
a\log\bigl[p(x)\sqrt{x^4+4x^3-6x^2+4x+1}+q(x)\bigr],
$$
where $a$ is a constant and $p$ and $q$ are polynomials. And indeed, after a painful differentiation and comparison, it turns out that the function
$$
\begin{aligned}
-\frac{1}{6}\log\Bigl[&\bigl(x^4+10x^3+30x^2+22x-11\bigr)\sqrt{x^4+4x^3-6x^2+4x+1}\\
&\qquad-x^6-12x^5-45x^4-44x^3+33x^2-43\Bigr]+C
\end{aligned}
$$
does the job.
