Here's an interesting integral that I'm struggling with.

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

MY EFFORTS: I'm very very close in reaching a closed form.

\begin{align} &= \frac12\int \frac{2\ dx}{\sqrt{\left(x^2+\dfrac{1}{x^2}\right) + 4\left(x+\dfrac{1}{x}\right)-6}} \\ &= \frac12\int \frac{\left(1 - \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x+\dfrac{1}{x}\right)^2 + 4\left(x+\dfrac{1}{x}\right)-8}} + \frac12\int \frac{\left(1 + \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x-\dfrac{1}{x}\right)^2 + 4\sqrt{\left(x-\dfrac{1}{x}\right)^2+4}-4}} \\ &= \frac12 \int \frac{du}{\sqrt{u^2+4u-8}} + \underbrace{\frac12 \int \frac{dt}{\sqrt{t^2 + 4\sqrt{t^2+4}-4}}}_{\text{How to solve this?}} \end{align}

As you all can see, I'm able to break integral by 2 parts and I'm struck in evaluating 2nd part. The first part can easily be solved by completing the square and trigonometric substitutions.

NOTE: In my effort pic. the variable of integration is x and should not be confused with letter n.

([1]: https://i.stack.imgur.com/k2Vhu.jpg)

  • $\begingroup$ Why the downvote? I've thinking to solve this integral for hours! Finally I'm just one step away from solving it. :( $\endgroup$ – Shivansh J Mar 14 '19 at 13:36
  • 1
    $\begingroup$ This integral leads to an elliptic one. $\endgroup$ – Dr. Sonnhard Graubner Mar 14 '19 at 13:36
  • $\begingroup$ What's an elliptical one integral? I'm sorry but I'm not much fimiliar with advanced maths. Can you please elaborate? $\endgroup$ – Shivansh J Mar 14 '19 at 13:37
  • $\begingroup$ @ShivanshJ en.wikipedia.org/wiki/Elliptic_integral $\endgroup$ – J.G. Mar 14 '19 at 13:40
  • $\begingroup$ You were downvoted because the question shows a lack of formatting. I've edited it for you, but for future reference here is a quick tutorial $\endgroup$ – Dylan Mar 14 '19 at 13:40

I don't think that your effort leads to a simpler integral that the original one. They are two radicals in your last integral. This makes it more difficult to integrate. So, you are not close to solve it.


$$x^4+4x^3-6x^2+x^4+1= (x-r_1)(x-r_2)(x-r_3)(x-r_4)$$ $r_1=\frac12\left(1+\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$

$r_2=\frac12\left(1-\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$

$r_3=\frac12\left(1+\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$

$r_4=\frac12\left(1-\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$

$$I= \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$ It is easier to integrate on this form : $$I= \int \frac{x\ dx}{\sqrt{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}} $$ The result involves an elliptic integral as a term of a complicated formula. From WolframAlpha :



If the integral comes from an academic exercise, there is probably a typo in it, because solving it requires an hight level of knowledge of special functions. I will not loose time with a problem suspected of mistake in the wording.

Please check your calculus which leads to the integral. Eventually edit the original version of the problem in your question.

  • $\begingroup$ You are right, it was a question given in academic exercise. But I don't know if it's really a typo or not because my instructor said that worksheet has tough questions of integration. Is the high level question fits for a 12th grader? $\endgroup$ – Shivansh J Mar 14 '19 at 15:36
  • $\begingroup$ Oh! But I'm pretty sad that all my process of 'solving' this problem was in vain. :( I wish I knew about special functions. $\endgroup$ – Shivansh J Mar 14 '19 at 15:38
  • $\begingroup$ @JJacquelin Just because WolframAlpha does not have an elementary closed antiderivative does not mean one does not exist, or that the original question contains a typo. A very simple example of an integrand for which WolframAlpha (and Mathematica) fails to compute an elementary closed form antiderivative is $$\int \frac{1-x^4}{(1+x^2+x^4)\sqrt{1 + x^4}} \, dx = \tan^{-1} \frac{x}{\sqrt{1+x^4}} + C.$$ $\endgroup$ – heropup Mar 14 '19 at 15:47
  • $\begingroup$ @heropup. My comment is not based on the WolframAlpha result (which I show for information only). It is based on the expressions of the roots $r_1,r_2,r_3,r_4$ which seems to not allow simplification. $\endgroup$ – JJacquelin Mar 14 '19 at 16:10
  • $\begingroup$ @Shivansh. Edit the original wording of your exercise into your question. (use the "edit" button. $\endgroup$ – JJacquelin Mar 14 '19 at 16:14

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