I came across this integral

$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$

According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.

A promising way might be the following

\begin{align*} \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\ &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3} \right )^2 + \frac{8}{3}}} \\ &= \cdots \end{align*}

A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )

So, any ideas how to evaluate it?

P.S: Are there techniques available for these type of problems?

  • 2
    $\begingroup$ WA doesn't give $\frac12$ but $0.500539$ $\endgroup$ – Raffaele Jan 21 at 21:15

According to Maple, the integral is $$\frac{\sqrt{3}\, \boldsymbol{\mathit{K}}\! \left(\frac{\sqrt{3}}{3}\right)}{6}$$ Due to different conventions, this would be $Sqrt[3] EllipticK[1/3]/6$ in Mathematica or Wolfram Alpha.


Since the integrand involves the square root of a quartic without repeated roots, elliptic integrals are needed. However, computer algebra systems usually give suboptimal results for such integrals – this is why I put together Elliptic Integrals and Functions so the "nicest" result can be obtained.

The corresponding Byrd and Friedman entry in this case is 225.00: $$\int_0^1\frac1{\sqrt{3x^4+2x^2+3}}\,dx=\frac1{\sqrt3}\int_0^1\frac1{\sqrt{(x^2+z^2)(x^2+\overline z^2)}}\,dx\qquad z=\sqrt{\frac23}+\frac1{\sqrt3}i$$ $$=\frac1{\sqrt3}\cdot\frac12F(2\tan^{-1}1,m=1/3)=\frac1{\sqrt{12}}K\left(\frac13\right)$$ The numerical value is most definitely not $\frac12$; it is $0.500538690228\dots$


This is an elliptic integral of the first kind, so there is no analytical way to evaluate it.

More on Elliptical Integrals: https://en.wikipedia.org/wiki/Elliptic_integral

In any case, we might try to give your integral a numerical result because it's bounded between $0$ and $1$.

We can expand the whole integrand in Taylor-MacLaurin series. Notice that the very first term of the expansion would just be $f(0) = \frac{1}{\sqrt{3}}$, so:

Order zero

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}}$$


$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} \approx 0.57735(...)$$

Let's go on with another term.

Second Order

Going on with the expansion we find:

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} + O(x^4)$$


$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\int_0^1 x^2\ \text{d}x = \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\cdot \frac{1}{3} = \frac{8}{9 \sqrt{3}} \approx 0.5132(...)$$

We can go on with the terms.

Numerical Exact Result

Via the help of W. Mathematica, the numerical inteegration gives the result


Which is rather near to our last result, as you can see.

Let's do better.

Fourth Order

$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} - \frac{x^4}{3 \sqrt{3}} + O\left(x^6\right)$$

So again

$$\int_0^1 f(x)\ \text{d}x \approx \frac{8}{9 \sqrt{3}} - \int_0^1 \frac{x^4}{3 \sqrt{3}} = \frac{37}{45 \sqrt{3}} \approx 0.47471(...) $$

It starts to oscillate around $0.500$. You can go on with the terms, what you will have to integrate are just polynomials.

A Big Order

Say we go on to order ten:

$$\frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}+O\left(x^{12}\right)$$


$$\int_0^1 f(x)\ \text{d}x \approx \int_0^1 \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}\ \text{d}x = \frac{238984}{280665 \sqrt{3}} \approx 0.491609(...)$$

Got it?

  • $\begingroup$ Wolfram Alpha doesn't give $1/2$ but $0.500539$ just like Mathematica. It would be weird if it was different :) $\endgroup$ – Raffaele Jan 21 at 21:14
  • $\begingroup$ @Raffaele Indeed, it's strange that Wolfram Alpha gives $1/2$ as an output... Perhaps it did approximate the solution (we all do know that W. Alpha online is severely bugged) $\endgroup$ – Turing Jan 21 at 22:36

A trigonometric substitution does indeed help here: Consider $f \colon [0,1] \to \mathbb{R},$ \begin{align} f (a) &= \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{1 + 2 a x^2 + x^4}} = \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{(1 + x^2)^2 - 2 (1-a) x^2}} \\ &\!\!\!\!\!\!\!\!\stackrel{x = \tan\left(\frac{t}{2}\right)}{=} \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - 2 (1-a) \sin^2 \left(\frac{t}{2}\right) \cos^2 \left(\frac{t}{2}\right)}} = \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - \frac{1-a}{2} \sin^2(t)}} \\ &= \frac{1}{2} \operatorname{K} \left(\sqrt{\frac{1-a}{2}}\right) \, . \end{align} Your integral is $$ \mathcal{J} = \frac{1}{\sqrt{3}} f \left(\frac{1}{3}\right) = \frac{\operatorname{K} \left(\frac{1}{\sqrt{3}}\right)}{2 \sqrt{3}} > \frac{1}{2} \, .$$


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