# An interesting indefinite integral

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.

• Why the downvote? I've thinking to solve this integral for hours! Finally I'm just one step away from solving it. :( – Shivansh J Mar 14 '19 at 13:36
• This integral leads to an elliptic one. – Dr. Sonnhard Graubner Mar 14 '19 at 13:36
• What's an elliptical one integral? I'm sorry but I'm not much fimiliar with advanced maths. Can you please elaborate? – Shivansh J Mar 14 '19 at 13:37
• @ShivanshJ en.wikipedia.org/wiki/Elliptic_integral – J.G. Mar 14 '19 at 13:40
• 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 – 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.

HINT :

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

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

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

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

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

https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt((x-r_1)(x-r_2)(x-r_3)(x-r_4))

NOTE :

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.

• @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 $$\int \frac{1-x^4}{(1+x^2+x^4)\sqrt{1 + x^4}} \, dx = \tan^{-1} \frac{x}{\sqrt{1+x^4}} + C.$$ – heropup Mar 14 '19 at 15:47
• @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. – JJacquelin Mar 14 '19 at 16:10