<s>[INST] $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
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.
 [/INST] 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:
$$\int\frac{x dx}{(x+1)^2}=\int\frac{(x+1) dx}{(x+1)^2}-\int\frac{1 dx}{(x+1)^2}=\ln(x+1)+\frac{1}{x+1}+C$$
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