On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$ 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?
 A: 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.
A: 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 $0.500538690228\dots$
A: 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
$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}}$$
Hence
$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} \approx 0.57735(...)$$
Let's go on with another term.
Second Order
Going on with the expansion we find:
$$f(x) = \frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \dfrac{1}{\sqrt{3}} - \frac{x^2}{3\sqrt{3}} + O(x^4)$$
Whence
$$\int_0^1 f(x)\ \text{d}x \approx \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\int_0^1 x^2\ \text{d}x = \dfrac{1}{\sqrt{3}} - \dfrac{1}{3\sqrt{3}}\cdot \frac{1}{3}  = \frac{8}{9 \sqrt{3}} \approx 0.5132(...)$$
We can go on with the terms.
Numerical Exact Result
Via the help of W. Mathematica, the numerical inteegration gives the result
$$0.500539(...)$$
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:
$$\frac{1}{\sqrt{3x^4 + 2x^2 + 3}} \approx \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}+O\left(x^{12}\right)$$
Whence
$$\int_0^1 f(x)\ \text{d}x \approx \int_0^1 \frac{1}{\sqrt{3}}-\frac{x^2}{3 \sqrt{3}}-\frac{x^4}{3 \sqrt{3}}+\frac{11 x^6}{27 \sqrt{3}}+\frac{x^8}{81 \sqrt{3}}-\frac{x^{10}}{3 \sqrt{3}}\ \text{d}x = \frac{238984}{280665 \sqrt{3}} \approx 0.491609(...)$$
Got it?
A: 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
$$ \mathcal{J} = \frac{1}{\sqrt{3}} f \left(\frac{1}{3}\right) = \frac{\operatorname{K} \left(\frac{1}{\sqrt{3}}\right)}{2 \sqrt{3}} > \frac{1}{2} \, .$$
