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$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$

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.

Brahmagupta
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    I have to wonder if there's a typo. Change a sign or 3 and you get a perfect square. – Paul Sep 21 '15 at 17:40
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    @paul Well, either you're right about there being a typo or the integral is elliptic, which means the integral is either very basic or very advanced. We can probably figure out which it is based on the level of the OP's course. – David H Sep 21 '15 at 17:52

3 Answers3

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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

Uri Goren
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  • Down voter, an explanation would be nice – Uri Goren Sep 23 '15 at 17:52
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    I like the approach, but after all it does not at all answer the question asked. – mickep Sep 23 '15 at 18:14
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    I didn't downvote (although I was tempted to). The integral you are left with looks at least as bad as the one you started with, so it's far from clear what the point of your manipulations are. – mrf Sep 23 '15 at 20:53
  • It's true, I didn't go all the way with this question, However, I suspect that that following my steps would yield an equation of the form $I=f(x)+g(x)I$ that would be solvable. – Uri Goren Sep 24 '15 at 07:28
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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.

mickep
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – vonbrand Sep 21 '15 at 19:02
  • Funny answer!!! – Daniel Sep 21 '15 at 19:15
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    @vonbrand I have now updated the answer and undeleted it, since I think this is the closest to a solution one will get (without doing the non-intuitive change of variables, setting $u=$ what is inside the logarithm). – mickep Sep 23 '15 at 16:01
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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: $$\int\frac{x dx}{(x+1)^2}=\int\frac{(x+1) dx}{(x+1)^2}-\int\frac{1 dx}{(x+1)^2}=\ln(x+1)+\frac{1}{x+1}+C$$

Uri Goren
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