$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
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.
 A: Note that:
$$(x+1)^4=x^4+4x^3+6x^2+4x+1$$
Then
$$(x+1)^4-12x^2=x^4+4x^3-6x^2+4x+1$$
$$(x+1)^4-12x^2=12x^2((\frac{(x+1)^2}{\sqrt{12}x})^2-1)$$
Let us define
$$f(x)=\frac{(x+1)^2}{\sqrt{12}x}$$
And derive it:
$$f'(x)=\frac{2(x+1)x-(x+1)^2}{12x^2}=\frac{x^2-1}{12x^2}$$
Now lets have a look at
$$[arcsin(f(x))]'=\frac{f'(x)}{\sqrt{f(^2x)-1}}=\frac{x^2-1}{12x^2\sqrt{(\frac{(x+1)^2}{\sqrt{12}x})^2-1}}=\frac{x^2-1}{12x\sqrt{x^4+4x^3-6x^2+4x+1}}$$
Simplifying a bit:
$$[arcsin(f(x))]'=\frac{x}{12\sqrt{x^4+4x^3-6x^2+4x+1}}-\frac{1}{12x\sqrt{x^4+4x^3-6x^2+4x+1}}$$
Thus:
$$\int{\frac{xdx}{\sqrt{x^4+4x^3-6x^2+4x+1}}}=12\arcsin(\frac{(x+1)^2}{\sqrt{12}x})+\int{\frac{dx}{x\sqrt{x^4+4x^3-6x^2+4x+1}}}$$
Now, that's not what you were looking for, but that's my best attempt
A: 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.
A: In case that there is a typo, as the comments suggest, and the function is:
$$\int\frac{x dx}{\sqrt{x^4+4x^3+6x^2+4x+1}}$$
Then the integral is fairly easy once noticing that:
$$x^4+4x^3+6x^2+4x+1=(x+1)^4$$
And thus, the integral can be simplified to:
$$\int\frac{x dx}{(x+1)^2}=\int\frac{(x+1) dx}{(x+1)^2}-\int\frac{1 dx}{(x+1)^2}=\ln(x+1)+\frac{1}{x+1}+C$$
