<s>[INST] 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?
 [/INST] 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?
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