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?

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?
