Integration $\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$ Here is the integral:
$$\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$$
This is an elliptic integral, with such an easy result, maybe some clever substitutions or integrating methods can solve it, but so far I don't know exactly how to attack it.
Thought 1:
To separate the integration interval into two parts:$(0,\frac12]$ and $[\frac12,+\infty)$, for the second one, doing a substitution $x\mapsto \frac1x$ is my first thought(possibly not valid).
Thought 2:
It looks 100%, even 90%, like pseudo elliptic integral(its anti-derivative is elementary). I think this one is elementary as well.
THOUGHT 3:
Maybe this is a very STUPID question. Perhaps some mathematical softwares can done it(especially for Mathematica). For some good reasons, I can't use them.
Waiting for your replies.
 A: I would like to solve it directly.
For $|t|$ large enough, let
$$
\tag{1}J\left(t\right)=\int_{0}^{\infty}\frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x/t^2)} }\text{d}x.
$$
Firstly, apply the substitution $y=xt$, and
$$
(1)={t^{2/3}}\int_{0}^{\infty}\frac{2y-t}{y^{2/3}\sqrt{(y+t)((y+t)^3-y)} }\text{d}y.
$$
Secondly, let $z=y^{1/3}$, and
$$
(1)=-3t^{2/3}\int_{0}^{\infty}\frac{t-2z^3}{\sqrt{(z^3+t)((z^3+t)^3-z^3)} }\text{d}z.
$$
Note that
$$
\frac{t-2z^3}{\sqrt{(z^3+t)((z^3+t)^3-z^3)} }
=\frac{1}{\sqrt{1+\left ( \frac{2z^3+z+2t}{z^3-z+t}  \right )^3}}\cdot\frac{\mathrm{d}}{\mathrm{d}z} \frac{2z^3+z+2t}{z^3-z+t}.
$$
Eventually, we apply the substitution
$$
s= \frac{2z^3+z+2t}{z^3-z+t}.
$$
Since $z\in[0,\infty)$, $s\in(2,2]$. Therefore $J(t)$ must be $0$. For complex $t$, these substitutions need to relate contour integration. But it's also good to show that $J(t)=0$.

One checks,
$$
\int_{\sqrt{5}-2}^{1} \frac{2x-1}{x^{2/3}\sqrt{(1-x^2)
(x^2+4x-1)} }\text{d}x=0.
$$

Actually, by performing substitution $ s=\frac{2z^a+z+2t}{z^a-z+t},a>1$ ($a$ no need to be an integer), we obtain
$$
\int_{0}^{\infty}\frac{(a-1)x-1}{x^{1-1/a}\sqrt{(x+1)\left((x+1)^3-tx^\frac3a\right)} }\text{d}x\equiv 0.
$$

*

*Setting $a=5,6,12$, then the following three integrals are derived.
$$
\int_{0}^{\infty}\frac{4x-1}{x^{4/5}\sqrt{(x+1)\left ( 
(x+1)^3+x^\frac35\right ) } }\text{d}x=0,\\
\int_{0}^{\infty}\frac{5x-1}{x^{5/6}\sqrt{(x+1)\left ( 
(x+1)^3+\sqrt{x}\right ) } }\text{d}x=0,\\
\int_{0}^{\infty}\frac{11x-1}{x^{11/12}\sqrt{(x+1)\left ( 
(x+1)^3+\sqrt[4]{x}\right ) } }\text{d}x=0.
$$

*For non-integer $a$: set $a=\frac{9}{2}$, and
$$
\int_{0}^{\infty}
\frac{7x^3-2}{x^{1/3}\sqrt{(x^3+1)\left ( (x^3+1)^3-t x^2 \right ) } } 
\text{d}x\equiv0.
$$
And we deduce the final one
$$
\int_{0}^{\infty}
\frac{(7x^3-2)f\left(\frac{tx^2}{(1+x^3)^3}\right)}{x^{4/3}\sqrt{1+x^3} } 
\text{d}x\equiv0.
$$
Where $f(x)$ could be almost any function that makes the integral convergent.

A: Let $$f(x,t) = \frac{2 x-1}{x^{2/3} \sqrt{(x+1) \left((x+1)^3-t x\right)}} \qquad I(t) = \int_0^\infty f(x,t) dx$$
with $g(x,t) = \frac{3 \left(x^2+x\right)}{2 x-1}f$, one checks
$$3t \frac{\partial f}{\partial t} + f+\frac{\partial g}{\partial x}= 0$$
so $$3tI'(t)+I(t) = -\int_0^\infty \frac{\partial g}{\partial x} dx = -g(\infty,t)+g(0,t) = 0$$
solving this ODE gives $I(t) = Ct^{-1/3}$. The function $t^{-1/3}$ blows up near $t=0$, but our $I(t)$ is continuous at $t=0$, this forces $C=0$. So $I(t)\equiv 0$ for $t$ near $0$. QED.
