Why does $|x-1| \; E\left(-\dfrac{4x}{(x-1)^2}\right) = |x+1| \; E\left(\dfrac{4x}{(x+1)^2}\right)$, where $E(m)$ is the complete elliptic integral of the second kind, with parameter $m$?
Answer
Note that the complete elliptic integral of the second kind satisfies the imaginary modulus identity (which I have specialized here to the complete case, $\phi=\pi/2$):
$$E(-m)=\sqrt{1+m}\,E\left(\frac{m}{1+m}\right)$$
Replacing $m$ with $\frac{4x}{(x-1)^2}$ in this identity gives
$$E\left(-\frac{4x}{(x-1)^2}\right)=\sqrt{1+\frac{4x}{(x-1)^2}}\;E\left(\frac{\frac{4x}{(x-1)^2}}{1+\frac{4x}{(x-1)^2}}\right)$$
which simplifies to
$$E\left(-\frac{4x}{(x-1)^2}\right)=\left|\frac{x+1}{x-1}\right|\;E\left(\frac{4x}{(x+1)^2}\right)$$
Multiplying both sides of the equation by $|x-1|$ turns this into what you have.
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